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Full text of "An elementary treatise on differential equations and their applications"

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http://www.archive.org/details/elementarytreatiOOpiaguoft 



BELL'S MATHEMATICAL SERIES 

ADVANCED SECTION ( £(■-* 

General Editor-. WILLIAM P. MILNE, M.A., D.Sc^ 
Professor of Mathematics, Leeds University 




AN ELEMENTARY TREATISE 

ON DIFFERENTIAL EQUATIONS AND 

THEIR APPLICATIONS 



y 



G. BELL AND SONS, LTD. 

LONDON : PORTUGAL ST., KINGSWAY 
CAMBRIDGE : DEIGHTON, BELL & CO. 
NEW YORK : THE MACMILLAN COM- 
PANY BOMBAY : A. H. WHEELER & CO. 



AN ELEMENTARY TREATISE ON 

DIFFERENTIAL 
EQUATIONS 

AND THEIR APPLICATIONS 



BY 



H. T. H. PIAGGIO, M.A., D.Sc. 

PROFESSOR OF MATHEMATICS, UNIVERSITY COLLEGE, NOTTINGHAM 
FORMERLY SENIOR SCHOLAR OF ST. JOHN'S COLLEGE, CAMBRIDGE 




\^ 



*jk\ 




V 



LONDON 

G. BELL AND SONS, LTD. 

1920 






V 



3D 

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Glasgow: printed at the university press 
by robert maclehose and co. ltd. 



PREFACE 

" The Theory of Differential Equations," said Sophus Lie, " is the 
most important branch of modern mathematics." The subject may 
be considered to occupy a central position from which different 
lines of development extend in many directions. If we travel along 
the purely analytical path, we are soon led to discuss Infinite Series, 
Existence Theorems and the Theory of Functions. Another leads 
us to the Differential Geometry of Curves and Surfaces. Between 
the two lies the path first discovered by Lie, leading to continuous 
groups of transformation and their geometrical interpretation. 
Diverging in another direction, we are led to the study of mechanical 
and electrical vibrations of all kinds and the important phenomenon 
of resonance. Certain partial differential equations form the start- 
ing point for the study of the conduction of heat, the transmission 
of electric waves, and many other branches of physics. Physical 
Chemistry, with its law of mass-action, is largely concerned with 
certain differential equations. 

The object of this book is to give an account of the central 
parts of the subject in as simple a form as possible, suitable for 
those with no previous knowledge of it, and yet at the same time 
to point out the different directions in which it may be developed. 
The greater part of the text and the examples in the body of it 
will be found very easy. The only previous knowledge assumed is 
that of the elements of the differential and integral calculus and a 
little coordinate geometry. The miscellaneous examples at the end 
of the various chapters are slightly harder. They contain several 
theorems of minor importance, with hints that should be sufficient 
to enable the student to solve them. They also contain geometrical 
and physical applications, but great care has been taken to state 
the questions in such a way that no knowledge of physics is required. 
For instance, one question asks for a solution of a certain partial 



vi PREFACE 

differential equation in terms of certain constants and variables. 
This may be regarded as a piece of pure mathematics, but it is 
immediately followed by a note pointing out that the work refers 
to a well-known experiment in heat, and giving the physical meaning 
of the constants and variables concerned. Finally, at the end of 
the book are given a set of 115 examples of much greater difficulty, 
most of which are taken from university examination papers. [I 
have to thank the Universities of London, Sheffield and Wales, and 
the Syndics of the Cambridge University Press for their kind per- 
mission in allowing me to use these.] The book covers the course 
in differential equations required for the London B.Sc. Honours or 
Schedule A of the Cambridge Mathematical Tripos, Part II., and 
also includes some of the work required for the London M.Sc. or 
Schedule B of the Mathematical Tripos. An appendix gives sugges- 
tions for further reading. The number of examples, both worked 
and un worked, is very large, and the answers to the un worked ones 
are given at the end of the book. 

A few special points may be mentioned. The graphical method 
in Chapter I. (based on the MS. kindly lent me by Dr. Brodetsky 
of a paper he read before the Mathematical Association, and on a 
somewhat similar paper by Prof. Takeo Wada) has not appeared 
before in any text-book. The chapter dealing with numerical 
integration deals with the subject rather more fully than usual. 
It is chiefly devoted to the methods of Runge and Picard, but it 
also gives an account of a new method due to the present writer. 

The chapter on linear differential equations with constant co- 
efficients avoids the unsatisfactory proofs involving " infinite con- 
stants." It also points out that the use of the operator D in finding 
particular integrals requires more justification than is usually given. 
The method here adopted is at first to use the operator boldly and 
obtain a result, and then to verify this result by direct differentiation. 

This chapter is followed immediately by one on Simple Partial 
Differential Equations (based on Riemann's " Partielle Differential - 
gleichungen "). The methods given are an obvious extension of 
those in the previous chapter, and they are of such great physical 
importance that it seems a pity to defer them until the later portions 
of the book, which is chiefly devoted to much more difficult subjects. 

In the sections dealing with Lagrange's linear partial differential 
equations, two examples have been taken from M. J. M. Hill's 
recent paper to illustrate his methods of obtaining special integrals. 



PREFACE vii 

In dealing with solution in series, great prominence has been 
given to the method of Frobenius. One chapter is devoted to the 
use of the method in working actual examples. This is followed 
by a much harder chapter, justifying the assumptions made and 
dealing with the difficult questions of convergence involved. An 
effort has been made to state very clearly and definitely where the 
difficulty lies, and what are the general ideas of the somewhat 
complicated proofs. It is a common experience that many students 
when first faced by a long " epsilon-proof " are so bewildered by 
the details that they have very little idea of the general trend. 
I have to thank Mr. S. Pollard, B.A., of Trinity College, Cambridge, 
for his valuable help with this chapter. This is the most advanced 
portion of the book, and, unlike the rest of it, requires a little know- 
ledge of infinite series. However, references to standard text-books 
have been given for every such theorem used. 

I have to thank Prof. W. P. Milne, the general editor of Bell's 
Mathematical Series, for his continual encouragement and criticism, 
and my colleagues Mr. J. Marshall, M.A., B.Sc, and Miss H. M. 
Browning, M.Sc, for their work in verifying the examples and 
drawing the diagrams. 

I shall be very grateful for any corrections or suggestions from 
those who use the book. 

H. T. H. PIAGGIO. 

University College, Nottingham, 
February, 1920. 



CONTENTS 



Historical Introduction 



PAOK 

XV 



CHAPTER I 

INTRODUCTION AND DEFINITIONS. ELIMINATION. 
GRAPHICAL REPRESENTATION 

ART. 

1-3. Introduction and definitions 1 

4-6. Formation of differential equations by elimination - - 2 
7-8. Complete Primitives, Particular Integrals, and Singular 

Solutions 4 

9. Brodetsky and Wada's method of graphical representation - 5 

10. Ordinary and Singular points 7 

Miscellaneous Examples on Chapter I - - - - 10 

CHAPTER II 

EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE 

11. Types to be considered 12 

12. Exact equations 12 

13. Integrating factors 13 

14. Variables separate ........ 13 

15-17. Homogeneous equations of the first order and degree - - 14 

18-21. Linear equations of the first order and degree - - - 16 

22. Geometrical problems. Orthogonal trajectories - - - 19 

Miscellaneous Examples on Chapter II - - - 22 



CHAPTER III 
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 



23. Type to be considered - 

24. Equations of the first order 



25 
25 



CONTENTS 

ART. 

25. Equations of the second order 

26. Modification when the auxiliary equation has imaginary or 

complex roots 

27. The case of equal roots -••---.. 

28. Extension to higher orders - 

29. The Complementary Function and the Particular Integral - 

30-33. Properties of the operator D 

34. Complementary Function when the auxiliary equation has 
repeated roots 

35-38. Symbolical methods of finding the Particular Integral. Ten- 
tative methods and the verification of the results they give 

39. The homogeneous linear equation 

40. Simultaneous linear equations ------ 

Miscellaneous Examples on Chapter III. (with notes on 
mechanical and electrical interpretations, free and 
forced vibrations and the phenomenon of resonance) 



CHAPTER IV 
SIMPLE PARTIAL DIFFERENTIAL EQUATIONS 

41. Physical origin of equations to be considered 

42-43. Elimination of arbitrary functions and constants - 

44. Special difficulties of partial differential equations - 

45-46. Particular solutions. Initial and boundary conditions - 

47-48. Fourier's Half-Range Series 

49-50. Application of Fourier's Series in forming solutions satisfying 
given boundary conditions ------ 

Miscellaneous Examples on Chapter IV. (with notes on 
the conduction of heat, the transmission of electric 
waves and the diffusion of dissolved salts) 



CHAPTER V 

EQUATIONS OF THE FIRST ORDER, BUT NOT OF THE 
FIRST DECREE 

51. Types to be considered -----... 

52. Equations solvable for p 

53. Equations solvable for y 

54. Equations solvable for x 



CONTENTS XI 

CHAPTER VI 
SINGULAR SOLUTIONS 

ART. PAGE 

55. The envelope gives a singular solution ----- 6"> 
56-58. The c-discriroinant contains the envelope (once), the node- 
locus (twice), and the cusp-locus (three times) 60 
59-64. The ^-discriminant contains the envelope (once), the tac-locus 

(twice), and the cusp-locus (once) - - - - - 71 

65. Examples of the identification of loci, using both discriminants 75 

66-67. Clairaut's form : - - 76 

Miscellaneous Examples on Chapter VI - - - - 78 

CHAPTER VII 

MISCELLANEOUS METHODS FOR EQUATIONS OF THE 
SECOND AND HIGHER ORDERS 

68. Types to be considered 81 

69-70. y or x absent 82 

71-73. Homogeneous equations 83 

74. An equation occurring in Dynamics 85 

75. Factorisation of the operator - - - - - - 86 

76-77. One integral belonging to the complementary function known 87 

78-80. Variation of Parameters 88 

81. Comparison of the different methods ----- 90 

Miscellaneous Examples on Chapter VII. (introducing the 
Normal form, the Invariant of an equation, and the 
Schwarzian Derivative) 91 

CHAPTER VIII 

NUMERICAL APPROXIMATIONS TO THE SOLUTION OF 
DIFFERENTIAL EQUATIONS 

82. Methods to be considered ------- 94 

83-84. Picard's method of integrating successive approximations - 94 

85. Numerical approximation direct from the differential equa- 
tion. Simple methods suggested by geometry 97 
86-87. Runge's method 99 

88. Extension to simultaneous equations 103 

89. Methods of Heun and Kutta 104 

90-93. Method of the present writer, with limits for the error - - 105 



xii CONTENTS 



CHAPTER IX 

SOLUTION IN SERIES. METHOD OF FROBENIUS 

ART. PAOK 

94. Frobenius' form of trial solution. The indicia] equation - 109 

95. Case I. Roots of indicial equation unequal and differing by a 

quantity not an integer - - 110 

96. Connection between the region of convergence of the series 

and the singularities of the coefficients in the differential 
equation 112 

97. Case II. Roots of indicial equation equal - - - - 112 

98. Case III. Roots of indicial equation differing by an integer, 

making a coefficient infinite - - - -. - - 114 

99. Case IV. Roots of indicial equation differing by an integer, 

making a coefficient indeterminate 116 

100. Some cases where the method fails. No regular integrals - 117 

Miscellaneous Examples on Chapter IX. (with notes on 
the hypergeometric series and its twenty-four solu- 
tions) 119 



CHAPTER X 

EXISTENCE THEOREMS OF PICARD, CAUCHY, AND 
FROBENIUS 

101. Nature of the problem 121 

102. Picard's method of successive approximation - - - 122 

103-105. Cauchy's method 124 

106-110. Frobenius' method. Differentiation of an infinite series with 

respect to a parameter - - - - - - -127 



CHAPTER XI 

ORDINARY DIFFERENTIAL EQUATIONS WITH THREE 
VARIABLES AND THE CORRESPONDING CURVES 
AND SURFACES 

111. The equations of this chapter express properties of curves and 

surfaces 133 

112. The simultaneous equations dx/P=dylQ=dz/R • - - 133 

113. Use of multipliers 135 

114. A second integral found by the help of the first - - - 136 

115. General and special integrals 137 



CONTENTS xiii 

ART. FAOB 

116. Geometrical interpretation of the equation 

Pdx+Qdy + Rdz=0 - - - - 137 

117. Method of integration of this equation when it is integrable - 138 

118-119. Necessary and sufficient condition that such an equation 

should be integrable 139 

120. Geometrical significance of the non-integrable equation - 142 

Miscellaneous Examples on Chapter XI - - - 143 



CHAPTER XII • 

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST 
ORDER. PARTICULAR METHODS 

121-122. Equations of this chapter of geometrical interest - - - 146 

123. Lagrange's linear equation and its geometrical interpretation 147 

124. Analytical verification of the general integral - - - 149 

125. Special integrals. Examples of M. J. M. Hill's methods of 

obtaining them 150 

126-127. The linear equation with n independent variables - - - 16*1 

128-129. Non-linear equations. Standard I. Only p and q present - 153 

130. Standard II. Only p, q, and z present 153 

131. Standard III. f(x, p)=F(y, q) 154 

132. Standard IV. Partial differential equations analogous to 

Clairaut's form 154 

133-135. Singular and General integrals and their geometrical signifi- 
cance. Characteristics 155 

136. Peculiarities of the linear equation 158 

Miscellaneous Examples on Chapter XII. (with a note on 

the Principle of Duality) 160 



CHAPTER XIII 

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST 
ORDER. GENERAL METHODS 

137. Methods to be discussed 162 

138-139. Charpit's method 162 

140-141. Three or more independent variables. Jacobi's method - 165 

142. Simultaneous partial differential equations - 168 

Miscellaneous Examples on Chapter XIII - - - 170 



xiv CONTENTS 



CHAPTER XIV 

PARTIAL DIFFERENTIAL EQUATIONS OF THE SECOND 
AND HIGHER ORDERS 

ART. PAQ 

143. Types to be considered 171] 

144. Equations that can be integrated by inspection. Determina- 

tion of arbitrary functions by geometrical conditions - 17! 

145-151. Linear partial differential equations with constant coefficients 171. 

152-153. Examples in elimination, introductory to Monge's methods - 17!> 

154. Monge's method of integrating Rr + Ss + Tt = V - - - 181 

155. Monge's method of integrating Rr + Ss + Tt + U(rt -s-) = V - 18.'! 
156-157. Formation of Intermediate Integrals 183 

158. Further integration of Intermediate Integrals - - - 186 

Miscellaneous Examples on Chapter XIV. (with notes on 
the vibrations of strings, bars, and membranes, and 
on potential) 188 

APPENDIX A 

Necessary and sufficient condition that the equation 
Mdx+Ndy=0 
should be exact - - 191 

APPENDIX B 

An equation with no special integrals ----- 192 

APPENDIX C 

The equation found by Jacobi's method .of Art. 140 is 

always integrable -----... 193 

APPENDIX D 

Suggestions for further reading -----.. 194 

Miscellaneous Examples <>n the Whole Book (with 
notes on solution by definite intregals, asymptotic series, 
the Wronskian, Jacobi's last multiplier, finite difference 
equations, Hamilton's dynamical equations, Foueault's 
pendulum, and the perihelion of Mercury) - - . 195 

Answers t<> the Examples j 

,S|JKX xxiii 



HISTORICAL INTRODUCTION 

The study of Differential Equations began very soon after the 
invention of the Differential and Integral Calculus, to which it 
forms a natural sequel. Newton in 1676 solved a differential 
equation by the use of an infinite series, only eleven years after 
his discovery of the fluxional form of the differential calculus in 
1665. But these results were not published until 1693, the same 
year in which a differential equation occurred for the first time in 
the work of Leibniz * (whose account of the differential calculus 
was published in 1684). 

In the next few years progress was rapid. In 1694-97 John 
Bernoulli f explained the method of " Separating the Variables," and 
he showed how to reduce a homogeneous differential equation of 
the first order to one in which the variables were separable. He 
applied these methods to problems on orthogonal trajectories. He 
and his brother Jacob ft (after whom " Bernoulli's Equation " is 
named) succeeded in reducing a large number of differential equa- 
tions to forms they could solve. Integrating Factors were probably 
discovered by Euler (1734) and (independently of him) by Fontaine 
and Clairaut, though some attribute them to Leibniz. Singular 
Solutions, noticed by Leibniz (1694) and Brook Taylor (1715), are 
generally associated with the name of Clairaut (1734). The geo- 
metrical interpretation was given by Lagrange in 1774, but the 
theory in its present form was not given until much later by Cayley 
(1872) and M. J. M. Hill (1888). 

The first methods of solving differential equations of the second 
or higher orders with constant coefficients were due to Euler. 
D'Alembert dealt with the case when the auxiliary equation had 
equal roots. Some of the symbolical methods of finding the par- 
ticular integral were not given until about a hundred years later 
by Lobatto (1837) and Boole (1859). 

The first partial differential equation to be noticed was that 
giving the form of a vibrating string. This equation, which is of 
the second order, was discussed by Euler and D'Alembert in 1747. 
Lagrange completed the solution of this equation, and he also 

* Also spelt Leibnitz. f Also spelt Bcrnouilli. ft Also known as James. 

xv 



XVI HISTORICAL INTRODUCTION 

dealt, in a series of memoirs from 1772 to 1785, with partial dif- 
ferential equations of the first order. He gave the general integral 
of the linear equation, and classified the different kinds of integrals 
possible when the equation is not linear. 

These theories still remain in an unfinished state ; contributions 
have been made recently by Chrystal (1892) and Hill (1917). Other 
methods for dealing with partial differential equations of the first 
order were given by Charpit (1784) and Jacobi (1836). For higher 
orders the most important investigations are those of Laplace (1773), 
Monge (1784), Ampere (1814), and Darboux (1870). 

By about 1800 the subject of differential equations in its original 
aspect, namely the solution in a form involving only a finite number 
of known functions (or their integrals), was in much the same state 
as it is to-day. At first mathematicians had hoped to solve every 
differential equation in this way, but their efforts proved as fruitless 
as those of mathematicians of an earlier date to solve the general 
algebraic equation of the fifth or higher degree. The subject now 
became transformed, becoming closely allied t<5 the Theory of 
Functions. Cauchy in 1823 proved that the infinite series obtained 
from a differential equation was convergent, and so really did 
define a function satisfying the equation. Questions of convergency 
(for which Cauchy was the first to give tests) are very prominent 
in all the investigations of this second period of the study of dif- 
ferential equations. Unfortunately this makes the subject very 
abstract and difficult for the student to grasp. In the first period 
the equations were not only simpler in themselves, but were studied 
in close connection with mechanics and physics, which indeed were 
often the starting point of the work. 

Cauchy's investigations were continued by Briot and Bouquet 
(1856), and a new method, that of " Successive Approximations," 
was introduced by Picard (1890). Fuchs (1866) and Frobenius 
(1873) have studied linear equations of the second and higher 
orders with variable coefficients. Lie's Theory of Continuous 
Groups (from 1884) has revealed a unity underlying apparently 
disconnected methods. Schwarz, Klein, and Goursat have made 
their work easier to grasp by the introduction of graphical con- 
siderations, and a recent paper by Wada (1917) has given a graphical 
representation of the results of Picard and Poincarr. Runge (1895) 
and others have dealt with numerical approximations. 

Further historical notes will be found in appropriate places 
throughout the book. For more detailed biographies, see Rouse 
Ball's Short History of Mathematics. 



CHAPTER I 



INTRODUCTION AND DEFINITIONS. ELIMINATION. 
GRAPHICAL REPRESENTATION 



^=-P 2 y, (i) 



1. Equations such as 

dx 2 

Miyf-g < 3 > 

d v = t u) 

dx y^l+x 1 )' 

dt 2 dx 2 ' {0) 

involving differential coefficients, are called Differential Equations. 

2. Differential Equations arise from many problems in Algebra, 
Geometry, Mechanics, Physics, and Chemistry. In various places 
in this book we shall give examples of these, including applications 
to elimination, tangency, curvature, envelopes, oscillations of 
mechanical systems and of electric currents, bending of beams, 
conduction of heat, diffusion of solvents, velocity of chemical 
reactions, etc. 

3. Definitions. Differential equations which involve only one 
independent variable,* like (1), (2), (3), and (4), are called ordinary. 

Those which involve two or more independent variables and 
partial differential coefficients with respect to them, such as (5), are 
called partial. 

* In equations (1), (2), (3), (4) x is the independent and y the dependent variable. 
In (5) x and ( are the two independent variables and y the dependent. 

p.d.e. a S 



2 DIFFERENTIAL EQUATIONS 

An equation like (1), which involves a second differential co- 
efficient, but none of higher orders, is said to be of the second order 
(4) is of the first order, (3) and (5) of the second; and (2) of the third. 

The degree of an equation is the degree of the highest differential 
coefficient when the equation has been made rational and integral 
as far as the differential coefficients are concerned. Thus (1), (2), 
(4) and (5) are of the first degree. 

(3) must be squared to rationalise it. We then see that it is of 

d 2 v 
the second degree, as j~ occurs squared. 

Notice that this definition of degree does not require x or y to 
occur rationally or integrally. 

Other definitions will be introduced when they are required. 

4. Formation of differential equations by elimination. The 

problem of elimination will now be considered, chiefly because it 
gives us an idea as to what kind of solution a differential equation 
may have. 

We shall give some examples of the elimination of arbitrary 
constants by the formation of ordinary differential equations. Later 
(Chap. IV.) we shall see that partial differential equations may be 
formed by the elimination of either arbitrary constants or arbitrary 
functions. 

5. Examples. 

(i) Consider x = A cos (pt- a), the equation of simple harmonic 
motion. Let us eliminate the arbitrary constants A and a. 

Differentiating, -=- = -pA sin (pt - a) 

d 2 x 
and -572 = - p 2 A cos (pt -a)= - p 2 x. 

d 2 x 
Thus -j-g = -p 2 x is the result required, an equation of the second 

order, whose interpretation is that the acceleration varies as the distance 
from the origin. 

(ii) Eliminate p from the last result. 

d x dx 

Differentiating again, -5-3 = - p 2 -57 • 

d?x \ dx d x \ 

Hence -373 -57 = - -p' J = ■■-,■■{ • x, (from the last result). 

3,'X Q.X d X 

Multiplying up, x . -=-3 » -j- • j-§, an equation of the third order. 



ELIMINATION 3 

(iii) Form the differential equation of all parabolas whose axis is 
the axis of x. 

Such a parabola must have an equation of the form 

y 2 = ia(x-h). 

Differentiating twice, we get 

i.e. y d £=2a, 

(L 11 i '(Lti\ 

and Vj\+ \y) = ^> wn ^ cn w °f the second order. 

Examples for solution. 

Eliminate the arbitrary constants from the following equations : 
(1) y = Ae 2x + Be~ 2x . (2) y = A cos 3x + B sin 3x. 

(3) y = Ae Bx . (4) y = Ax + A*. 

(5) If x 2 + y 2 = a 2 ,^ prove that j-= --, and interpret the result 
geometrically. & 

(6) Prove that for any straight line through the origin -»^, and 
interpret this. * dx 

d 2 u 

(7) Prove that for any straight line whatever -r\ = 0. Interpret 

this. dx 

6. To eliminate n arbitrary constants requires (in general) a differ- 
ential equation of the n^ order. The reader will probably have 
arrived at this conclusion already, from the examples of Art. 5. 
If we differentiate n times an equation containing n arbitrary con- 
stants, we shall obtain (n + 1) equations altogether, from which the 
n constants can be eliminated. As the result contains an n th differ- 
ential coefficient, it is of the n th order.* 

* The ?'-^ument in the text is that usually given, but the advanced student 
will notice some weak points in it. The statement that from any (n + 1) equations 
n quantities can be eliminated, whatever the nature of those equations, is too sweeping. 
An exact statement of the necessary and sufficient conditions would be extremely 
complicated. 

Sometimes less than (n + 1) equations are required. An obvious case is 
y = (A + B)x, where the two arbitrary constants occur in such a way as to be 
really equivalent to one. 

A less obvious case is y 2 =2Axy + Bx 2 . This represents two straight lines 
through the origin, say y = m l x and y=m 2 x, from each of which we easily get 

-= j-, of the first instead of the second order. The student should also obtain 
x dx 

this result by differentiating the original equation and eliminating B. This will 

give . , . 

[y-x£)(y-Ax) = 0. 



4 DIFFERENTIAL EQUATIONS 

7. The most general solution of an ordinary differential equation of 
the n th order contains n arbitrary constants. This will probably seem 
obvious from the converse theorem that in general n arbitrary con- 
stants can be eliminated by a differential equation of the n th order. 
But a rigorous proof offers much difficulty. 

If, however, we assume * that a differential equation has a solution 
expansible in a convergent series of ascending integral powers of 
x, we can easily see why the arbitrary constants are n in number. 

Consider, for example, ^-| = ^, of order three. 

Assume that y = a + a 1 x+a 2 ^ + ... + a M — f + ... to infinity. 
Then, substituting in the differential equation, we get 

X 2 X n ~ 3 ... x 2 x n ~ x 

so a 3 =a v 

a i = a 2 , 

tt 5 =a 3 =0! l> 

/ X 3 X 5 \ (x 2 X* T 6 \ 

Hence y-*+^+ S +fl + "0+<2! + II + fl + "0 

= a Q +a i sinh x + a 2 (cosh x - 1), 
containing three arbitrary constants, a , a x and a 2 . 
Similar reasoning applies to the equation 

^y =f ( r „ <k ^i dn ~ l y \ 
dx n j\ x > y> dx > dx v ••> dx n~ij- 

In Dynamics the differential equations are usually of the second 

order, e.g. -j-f +p 2 y=0, the equation of simple harmonic motion. 

To get a solution without arbitrary constants we need Iwo con- 
ditions, such as the value of y and dyjdt when t = 0, giving the initial 
displacement and velocity. 

8. Complete Primitive, Particular Integral, Singular Solution. The 
solution of a differential equation containing the full number of 
arbitrary constants is called the Complete Primitive. 

Any solution derived from the Complete Primitive by giving 
particular values to these constants is called a Particular Integral. 

* The student will see in later chapters that this assumption is not always 
justifiable. 



GRAPHICAL REPRESENTATION 



Thus the Complete Primitive of -t4=-j 
r dx 3 dx 



is y = a Q +a 1 sinh x+a 2 (cosh x - 1), 

or ?/ = c + «j sinh x+a 2 cosh a;, where c = a Q - a 2 , 

or y=c+ae a! +&e~ a; , where a = l(a 1 +a 2 ) and 6 = |(a 2 -«i)- 

This illustrates the fact that the Complete Primitive may often 
be written in several different (but really equivalent) ways. 
The following are Particular Integrals : , 

y=4, taking c = 4, a 1 =a 2 =0; 

y = 5smhx, taking 0^=5, c = a 2 =0; 
y = 6 cosh z - 4, taking a 2 = 6, a x =0, c = - 4 ; 
y = 2+e x -3e~ x , taking c = 2, a = l, 6= -3. 
In most equations every solution can be derived from the Com- 
plete Primitive by giving suitable values to the arbitrary constants. 
Bowever, in some exceptional cases we shall find a solution, called 
a Singular Solution, that cannot be derived in this way. These will 
be discussed in Chap. VI. 

Examples for solution. 

Solve by the method of Art. 7 : 

« ■£* • 

« 3--* 

(3) Show that the method fails for ■£-— -. 

x ' ax x 

[log as cannot be expanded in a Maclaurin series.] 

(4) Verify by elimination of c that y = ca; + - is the Complete Primitive 

of v = x -T- + 1 / -^ . Verify also that y 2 = ix is a solution of the differential 
y dx I dx J * 

equation not derivable from the Complete Primitive {i.e. a Singular 

Solution). Show that the Singular Solution is the envelope of the 

family of lines represented by the Complete Primitive. Illustrate by 

a graph. 

9. Graphical representation. We shall now give some examples 
of a method * of sketching rapidly the general form of the family of 
curves representing the Complete Primitive of 

* Duo to Dr. S. Brodetsky and Prof. Takeo Wada. 



6 



DIFFERENTIAL EQUATIONS 



where f{x, y) is a function of x and y having a perfectly definite 
finite value * for every pair of finite values of x and y. 

The curves of the family are called the characteristics of the 
equation. , 



Ex. (i) 



Here 






Now a curve has its concavity upwards when the second differential 
coefficient is positive. Hence the characteristics will be concave up 
above y = \, and concave down below this line. The maximum or 
minimum points lie on x=0, since dy/dx = there. The characteristics 
near y = l, which is a member of the family, are flatter than those 
further from it. 

These considerations show us that the family ie of the general form 
shown in Fig. 1. 

y 




Fig. 1 



Ex. (ii) 



Here 



dy 

d 2 y dy 

-~=^r + e x = y + 2e a 

dx 2 dx 



We start by tracing the curve of maxima and minima y + e* = 0, 
and the curve of inflexions y + 2e x — 0. Consider the characteristic 
through the origin. At this point both differential coefficients are 
positive, so as x increases y increases also, and the curve is concave 
upwards. This gives us the right-hand portion of the characteristic 
marked 3 in Fig. 2. If we move to the left along this we get to the 

•Thus excluding a function hke yjx, which is indeterminate when a;=0 and 



GRAPHICAL REPRESENTATION 



curve of minima. At the point of intersection the tangent is parallel to 
Ox. After this we ascend again, so meeting the curve of inflexions. 
After crossing this the characteristic becomes convex upwards. It still 
ascends. Now the figure shows that if it cut the curve of minima again 

y 




Fig. 2. 



the tangent could not be parallel to Ox, so it cannot cut it at all, but 
becomes asymptotic to it. 

The other characteristics are of similar nature. 



Examples for solution. 






Sketch the characteristics of 


: 


(1) 


dx 


y{\-x). 


(2) 


dx 


x 2 y. 


(3) 


dy = 
dx 


y+x 2 . 



10. Singular points. In all examples like those in the last 
article, we get one characteristic, and only one, through every point 

dv d 2 v 

of the plane. By tracing the two curves -g- =0 and j\ =0 we can 

easily sketch the system. 

If, however, f(x, y) becomes indeterminate for one or more 

points (called singular points), it is often very difficult to sketch the 



8 DIFFERENTIAL EQUATIONS 

system in the neighbourhood of these points. But the following 
examples can be treated geometrically. In general, a complicated 
analytical treatment is required.* 



Ex. (i). 



■4-=-- Here the origin is a singular point. The geo- 

CLOO X 

metrical meaning of the equation is that the radius vector and the 
tangent have the same gradient, which can only be the case for straight 




Fig. 3. 

lines through the origin. As the number of these is infinite, in this case 
an infinite number of characteristics pass through the singular point. 

Ex.(ii). *--?, i.e. *.*--l. 

ax y x Gfe 

This means that the radius vector and the tangent have gradients 




whose product is -1, i.e. that they are perpendicular. The char- 
acteristics are therefore circles of any radius with the origin as centre. 

* See a paper, " Graphical Solution," by Prof. Takeo Wada, Memoirs of the 
College of Science, Kyoto Imperial University, Vol. II. No. 3, July 1917. 



GRAPHICAL REPRESENTATION 9 

In this case the singular point may be regarded as a circle of zero radius, 
the limiting form of the characteristics near it, but no characteristic of 
finite size passes through it. 

Bx.(iH). P'*^- 

v ax x + ky 

Writing dy/dx=>ta,n\fs, y/x = tan 6, we get 

, tan 0-k 
tan ^ = l + fctan0 ' 

i.e. tan \f, + k tan \\r tan = tan 6-k, 
tan - tan \[s 



i.e. 



k, 



1 + tan 6 tan \/r 

i.e. tan (d-\fs) = k, a constant. 

The characteristics are therefore equiangular spirals, of which the 
singular point (the origin) is the focus. 




Fig. 5. 

These three simple examples illustrate three typical cases. 

Sometimes a finite number of characteristics pass through a singular 

point, but an example of this would be too complicated to give 

here.* 

* See Wada's paper. 




10 DIFFERENTIAL EQUATIONS 



MISCELLANEOUS EXAMPLES ON CHAPTER I. 

Eliminate the arbitrary constants from the following : 

(1) y = Ae x + Ber x + C. 

(2) y = Ae x + Be 2x + C<? x . 

[To eliminate A, B,C from the four equations obtained by successive 
differentiation a determinant may be used.] 

(3) y = e x (A cos x + B sin x), 

(4) y = c cosh -, (the catenary). 

c 

Find the differential equation of 

(5) All parabolas whose axes are parallel to the axis of y. 

(6) All circles of radius a. 

(7) All circles that pass through the origin. 

(8) All circles (whatever their radii or positions in the plane xOy). 
[The result of Ex. 6 may be used.] 

(9) Show that the results of eliminating a from 

2y=xd £ +ax> (1) 

) dy 

and b from y = x-j--bx 2 , (2) 

d y dy 
are in each case x 2 ^A,-2x^- + 2y = (3) 

dx 2 dx J v ' 

[The complete primitive of equation (1) must satisfy equation (3), 
since (3) is derivable from (1). This primitive will contain a and also 
an arbitrary constant. Thus it is a solution of (3) containing two 
constants, both of which are arbitrary as far as (3) is concerned, as a 
does not occur in that equation. In fact, it must be the complete 
primitive of (3). Similarly the complete primitives of (2) and (3) are 
the same. Thus (1) and (2) have a common complete primitive.] 

(10) Apply the method of the last example to prove that 

y+y = 2ae x 
dx 

■and y-^~ = 2be- x 

* dx 

have a common complete primitive. 

(11) Assuming that the first two equations of Ex. 9 have a common 

dlJ 

complete primitive, find it by equating the two values of ~ in terms 

of x, y, and the constants. Verify that it satisfies equation (3) of Ex. 9. 

(12) Similarly obtain the common complete primitive of the two 
equations of Ex. 10. 



MISCELLANEOUS EXAMPLES 11 

(13) Prove that all curves satisfying the differential equation 

ax \dx/ ax* 

cut the axis of y at 45°. 

(14) Find the inclination to the axis of x at the point (1, 2) of the 
two curves which pass through that point and satisfy 

(|)V-2z + *s. 

(15) Prove that the radius of curvature of either of the curves of 
Ex. 14 at the point (1, 2) is 4. 

(16) Prove that in general two curves satisfying the differential 
equation 



•0EN2+i-« 



pass through any point, but that these coincide for any point on a 
certain parabola, which is the envelope of the curves of the system. 

(17) Find the locus of a point such that the two curves through it 
satisfying the differential equation of Ex. (16) cut (i) orthogonally ; 
(ii) at 45°. 

(18) Sketch (by Brodetsky and Wada's method) the characteristics of 

ax 



CHAPTER II 
EQUATIONS OF THE FIRST ORDER AND FIRST DEGREE 

11. In this chapter we shall consider equations of the form 

M+N^=0, 
ax 

where M and N are functions of both x and y. 

This equation is often written,* more symmetrically, as 
Mdx+Ndy=0. 

Unfortunately it is not possible to solve the general equation of 
this form in terms of a finite number of known functions, but we 
shall discuss some special types in which this can be done. 

It is usual to classify these types as 

(a) Exact equations ; • 

(b) Equations solvable by separation of the variables ; - 

(c) Homogeneous equations ; • 

(d) Linear equations of the first order. • 

The methods of this chapter are chiefly due to John Bernouilli 
of Bale (1667-1748), the most inspiring teacher of his time, and to 
his pupil, Leonhard Euler, also of Bale (1707-1783). Euler made 
great contributions to algebra, trigonometry, calculus, rigid dynamics, 
hydrodynamics, astronomy and other subjects. 

12. Exact equations, f 

Ex. (i). The expression ydx + xdy is an exact differential. 

Thus the equation ydx + xdy = 0, 

giving d{yx)=0, 

i.e. yx = c, 
is called an exact equation. 

* For a rigorous justification of the use of the differentials dx and dy see Hardy's 
Pure Mathematics, Art. 136. 

t For the necessary and sufficient condition that Mdx + Ndy = should be exact 
see Appendix A. 

12 



EQUATIONS OF FIRST ORDER AND FIRST DEGREE 13 

Ex. (ii). Consider the equation tan y . tfo + tan x . dy = 0. 

This is not exact as it stands, but if we multiply by cos x cos y it 
becomes sin y CO s x dx + sin x cos y dy = 0, 

which is exact. 

The solution is sin y sin x = c. y 

13. Integrating factors. In the last example cos x cos y is 
called an integrating factor, because when the equation is multiplied 
by it we get an exact equation which can be at once integrated. 

There are several rules which are usually given for determining 
integrating factors in particular classes of equations. These will be 
found in the miscellaneous examples at the end of the chapter. The 
proof of these rules forms an interesting exercise, but it is generally 
easier to solve examples without them. 

14. Variables separate. 

dx 
Ex. (i). In the equation — =tan y . dy, the left-hand side involves 

x only and the right-hand side y only, so the variables are separate. 
Integrating, we get log x = - log cos y + c, 
i.e. log (x cos y) = c, 
x cos y = e c = a, say. 

Ex. (ii). | = 2x*/. 

The variables are not separate at present, but they can easily be 
made so. Multiply by dx and divide by y. We get 

— =2xdx. 

V 
Integrating, log y = x 2 + c. 

As c is arbitrary, we may put it equal to log a, where a is another 
arbitrary constant. 

Thus, finally, y = ae x2 . 

Examples for solution. 

(1) (12x + 5y-9)dx + (5x + 2y-4)dy = 0. 

(2) {cos x tan y + cos (x + y)} dx.+ {sin x sec 2 y + cos (x + y)} dy = 0. 

(3) (sec x tan x tan y - e x ) dx + sec x sec 2 y dy -■= 0. 
* (4) (x + y) (dx - dy) =dx + dy. 
. (5) ydx-xdy + Sx^e^dx = 0. 

(6) y dx - x dy = 0. 
• " (7) (sin x + cos x) dy + (cos x - sin x) dx = 0. 

•<8) g=*y. 

,{9) y dx-x dy = xy dx. (10) tan x dy = cot y dx. 



14 DIFFERENTIAL EQUATIONS 

15. Homogeneous equations. A homogeneous equation of the 
first order and degree is one which can be written in the form 

dy=f(y\ 

dx J \x/ 
To test whether a function of x and y can be written in the form 
of the right-hand side, it is convenient to put 

y 

-=v or y=vx. 

00 

If the result is of the formf(v), i.e. if the x's all cancel, the 
test is satisfiM. 

-, ... dy x 2 + y 2 , dy \+v 2 m , . -. . , 

Ex. (i). j^= g becomes -f-= . This equation is homo- 

geneous. dx Zx dx l 

diJ w da 

Ex. (ii). ^=^2 becomes -^- = xv z . This is not homogeneous. 

CLOO 00 (LOO — 

16. Method of solution. Since a homogeneous equation can be 

dy 
reduced to ^f-=f( v ) by putting y=vx on the right-hand side, it is 

natural to try the effect of this substitution on the left-hand side 
also. As a matter of fact, it will be found that the equation can 
always be solved * by this substitution (see Ex. 10 of the miscel- 
laneous set at the end of this chapter). 

Ex. (i). f^ = ^. 

w dx 2x 2 

Put y=vx, 

i.e. -jr**v+z-z-> i for if y is a function of x, so is v). 
dx dx v 9 ' 

_,, . dv 1 + v 2 

The equation becomes v + x-z- = — ~ — , 



Separating the variables, 



i.e. 2x dv = (1 + v 2 - 2v) dx. 
2dv dx 



(v-l) 2 x 



— 2 

Integrating, — - = logx + c. 

y -2 _ -2 - 2x _ 2x 

u ' v ~x' ° v z i~y_ 1 ~y-x~x-y' 

X 

Multiplying by x-y, 2x = (x - y) (log x + c) . 

* By " solved " we mean reduced to an ordinary integration. Of course, this 
integral may not be expressible in terms of ordinary elementary functions. 



EQUATIONS OF FIRST ORDER AND FIRST DEGREE 15 

Ex. (ii). (x + y) dy + (x - y) dx = 0. 

m ,. . dy y-x 

This gives ~ = - — • 

dx y + x 

Putting y = vx, and proceeding as before, we get 

dv v-1 

dx v + l 

dv v-l v 2 + l 

i.e. x^-= — --« = -. 

dx v + l v + l 

(v + l)dv dx 



Separating the variables, - 



i.e. 



v 2 + l x 
v dv dv dx 



v 2 + l v 2 + l X 
Integrating, - £ log (v 2 + 1) - tan -1 v - log x + c, 

i.e. 2 log x + log (v 2 + 1) + 2 tan -1 ?; + 2c = 0, 
logx 2 ( , y 2 + l)+2tan~M+a = 0, putting 2c = a. 

Substituting for v, log {y 2 + x 2 ) + 2 tan -1 - + a = 0. 

•2/ 

17. Equations reducible to the homogeneous form. 

-n /•* mi ^ dy y-x + \, 

Ex. (i). The equation ■#=- = 

u dx y + x + b 

is not homogeneous. 

This example is similar to Ex. (ii) of the last article, except that 

y-x . , -, i y-x + \ 

is replaced by -. 

y+x r J y+x+5 

Now y-x=0 and y + x = represent two straight lines through the 
origin. 

The intersection of y-x+l=0 and y + x + 5 = is easily found to 
be (-2, -3). 

Put x = X - 2 ; y=Y -3. This amounts to taking new axes parallel 
to the old with ( - 2, - 3) as the new origin. 

Then y-x + l = Y-X and y + x + 5=Y + X. 

Also dx = dX and dy = dY. 

The equation becomes -t^> = -^ — =>• 
* dX Y + X 

As in the last article, the solution is 

log(Y 2 + X 2 )+2tan- 1 ^ + a = 0, 
A 

i.e. log[(2/ + 3) 2 + (z + 2) 2 ] + 2tan- 1 ^ + a = 0. 

x + A 



16 DIFFERENTIAL EQUATIONS 

Ex. (u). /=^ -. 

dx y-x + o 

This equation cannot be treated as the last example, because the 
lines y-x+\=0 and y-x + 5 = are parallel. 

As the right-hand side is a function oiy-x, try putting y-x = z, 

dy _dz 
dx dx' 

The equation becomes 1 + ^- = - — -, 

' m z + 5 

dz -4 
i.e. -j- = — =. 
dx z + 5 

Separating the variables, (z + 5) dz = - 4 dx. 

Integrating, |z 2 + 5z = - 4sc + c, 

i.e. z 2 + 10z + 8x = 2c. 

Substituting for z, (y - x) 2 + 10 (y - x) + 8x = 2c, 

i.e. (y-x) 2 + \0y -2x = a, putting 2c = a. 

Examples for solution. 

•(1) {2x-y)dy = (2y-x)dx. [Wales.] 

(2) {x*-y 2 )^- = xy. [Sheffield.] 

^- (3)2 tH + & [MatL Tripos - ] 

.(4) xf x = y + V(x* + y 2 )- 

dy_ 2x + 9y-20 
1 ' dx 6x + 2y-10* 
(6) (12» + 21y-9)flte + (47a? + 40y + 7)dy-0. 

> dy_ 3x-iy-2 
[ ' dx 3x-4y-3" 

(8) (jB + 2y)(ia5-dy)=*B + dy. 
18. Linear equations. 
The equation ~£ +Py = ®' 

where P and Q are functions of x (but not of y), is said to be linear 

of the first order. 

', . dy 1 „ 

A simple example is -^ + - . y =x 2 . 

{ 



EQUATIONS OF FIRST ORDER AND FIRST DEGREE 17 

If we multiply each side of this by x, it becomes 



x ix + y =x > 



ie ' dx^ =X ^ 

Hence, integrating, xy = \x* + c. 

We have solved this example by the use of the obvious integrating 
factor x. 

19. Let us try to find an integrating factor in the general case. 
If R is such a factor, then the left-hand side of 

Rf x + RPy = RQ 

is the differential coefficient of some product, and the first term 

R -j- shows that the product must be Ry. 

Put, therefore, R^+RPy=-^(Ry) =#^ + y~. 

This gives RPy = y-^, 

i.e. Pdx = -n, 
li 



i.e. 



ipdx=logR, 

[Pdx 



This gives the rule : To solve -j- +Py = Q, multiply each side by 

\pdx 

e , which will be an integrating factor. 

20. Examples. 

(i) Take the example considered in Art. 18. 

ty . 1 2 

-T-+- .y=<c 2 . 

ax x. 
HereP = -, so \Pdx = logx, and e lo « r =x.. 
Thus the rule gives the same integrating factor that we used before. 

(ii) g + 2a*/ = 2<r*\ 

Here P = 2x, \Pdx = x 2 , and the integrating factor is e**. 

P.D.E. B 

:- 



18 DIFFERENTIAL EQUATIONS 

Multiplying by this, e* 2 ~- + 2xe x% y = 2, 

Integrating, ye x * = 2x + c, 

y = (2x + c)e-° fi . 

(is, t + %-* 2 * 

Here the integrating factor is e 3x . 

Multiplying by this, e 3 * ^ + ^ x y ~ & x > 

a 
i.e. j-(ye Zx )-e 5x . 

Integrating, ye 3 x = -te 5 x + c, 

y = -Lg 2 * + ce -3 *. 

21. Equations reducible to the linear form. 

Ex. (i). xy-^-yh-*. 

Divide by y z , so as to free the right-hand side from y. 

ttT 1 1 dy ^ 

We get . tjS . jg.^, 



1 1 <Z /l 



$-*• 



Putting ^ = 2» 2a» + -=- = 2e~- 



i/ 2 2 cfoVy 5 
1 <fc 

This is linear and, in fact, is similar to Ex. (ii) of the last article with 
2 instead of y. 

Hence the solution is z = (2x + c) e~ x \ 

i.e. — = {2x + c)e~ x \ 

y 2 

ei* 2 



'J{2x+o) 

This example is a particular case of " Bernoulli's Equation " 

where P and Q are functions of x. Jacob Bernouilli or Bernoulli ol 
Bale (1654-1705) studied it inl695. 



EQUATIONS OF FIRST ORDER AND FIRST DEGREE 19 



Ex. (ii). 



(2*-i(y) d £+y=o. 



fix 



This is not linear as it stands, but if we multiply by -=-, we get 

2z-i(y+</|=o, • 

dx 2x ,. „ 
i.e. -T-+— = 10w 2 . 
dy y 

This is linear, considering y as the independent variable. 

Proceeding as before, we find the integrating factor to be y 2 , and 

the solution 2 o *, , 

y 2 x = 2y 5 + c, , 

i.e. a: = 2?/ 3 + c?/"- 2 . 
Examples for solution. 

/(I) (* + a)j|-8y-(a> + a)«. [Wales.] 

% (2) a;cosa;^ + «/(a;sincc + cosaj) = l. [Sheffield.] 

•(3) xloga?^ + «/ = 21ogcc. (4) x 2 y - x* j- = y* cos x. 



-7(5) y + 2f x = f(x-l). 



.(6) (* + 2jfl|[-y. 



• (7) dx + xdy = e~v sec 2 y dy. 

22. Geometrical Problems. Orthogonal Trajectories. We shall 
now consider some geometrical problems leading to differential 
equations. 

y 




Ex. (i). Find the curve whose subtangent is constant. 
The sttbtangent TN = PN cot yj, = y ~ . 



20 DIFFERENTIAL EQUATIONS 

Hence y -y- - k, 

*dy 

dx = k—, 

y 

x + c = k log y, 

putting the arbitrary constant c equal to k log a. 

Ex. (ii). Find the curve such that its length between any two 
points PQ is proportional to the ratio of the distances of Q and P 
from a fixed point 0. 

If we keep P fixed, the arc QP will vary as OQ. 

Use polar co-ordinates, taking as pole and OP as initial line. 
Then, if Q be (r, 6), we have s — ^ r> 

But, as shown in treatises on the Calculus, 
(ds)* = (rdd) z + (dr) 2 . 
Hence, in our problem, 

k* (dr) 2 = (rd6) 2 + (dr)*, 

i.e. <Z0=±V(& 2 -1)- 

ldr 

= , say, 

a r 

giving r = ce a0 , the equiangular spiral. 

Ex. (iii). Find the Orthogonal Trajectories of the family of semi- 
cubical parabolas ay 2 = x 3 , where a is a variable parameter. 

Two families of curves are said to be orthogonal trajectories when 
every member of one family cuts every member of the other at right 
angles. 

We first obtain the differential equation of the given family by 
eliminating a. 

Differentiating ay 2 = x 3 , 

we get 2ay -^- = 3x 2 , 

whence, by division, ~:r = - (1) 

' J y dx x 

Now ^=tan \ls, where \lr is the inclination of the tangent to the 
dx T 

axis of x. The value of \\r for the trajectory, say rfs', is given by 

\fr = \Js' ± \ir, 
i.e. tan \js= -cot yj/', 

i e. — for the given family is to be replaced by - -j- for the trajectory. 
dx ay 






EQUATIONS OF FIRST ORDER AND FIRST DEGREE 21 

Making this change in (1), we get 

_2dx 3 

ydy = x 

2xdx + 3y dy = 0, 

2x 2 + 3y 2 = c, 

a family of similar and similarly situated ellipses. 

Ex. (iv). Find the family of curves that cut the family of spirals 
r = a9 at a constant angle a. 

As before, we start by eliminating a. 

This gives -j- = 6. 

Now — j- =tan <f>, where is the angle between the tangent and the 

radius vector. If (/>' is the corresponding angle for the second family, 

0' = 0±a, 

, tan <j> ± tan a 6 + k 

tan <A =- — ~- — ■ = - — r7 :, 

r l+tan0tana 1 - kQ 

putting in the value found for tan <f> and writing k instead of ±tan a. 
Thus, for the second family, 

rdd^ 6 + k 
dr ~l-kO' 

The solution of this will be left as an exercise for the student. 
The result will be found to be 

r = c(6 + k) kl + 1 e- k9 . 



Examples for solution. 

(1) Find the curve whose subnormal is constant. 

(2) The tangent at any point P of a curve meets the axis of x in T. 
Find the curve for which OP — PT, being the origin. 

-(3) Find the curve for which the angle between the tangent and 
radius vector at any point is twice the vectorial angle. 

(4) Find the curve for which the projection of the ordinate on the 
normal is constant. 

Find the orthogonal trajectories of the following families of curves : 

(5) x 2 -y 2 = a 2 . • .(6) x$ + y* = a$. 
(7) px 2 + qy 2 = a 2 , (p and q constant). 

a6 



.(8) rd = a. (9) r = 



l+<9* 



(10) Find the family of curves that cut a family of concentric circles 
at a constant angle a. 



22 DIFFERENTIAL EQUATIONS 

MISCELLANEOUS EXAMPLES ON CHAPTER II. 

(I) (3y*-x) d £ = y. (2) * d £ = y + 2V(y 2 -x*). 

(3) tan x cos y dy + sin y dx + e 8 ' 11 x dx =0. 

(4) x*ijt + Zy* = xyK [Sheffield.] 

.(5) tff x =y z +yW(y 2 -x*)> 

(6) Show that 4- -+*+{ 
x ' da; hx + by+f 

represents a family of conies. 

(7) Show that ydx-2xdy = 

represents a system of parabolas with a common axis and tangent at 
the vertex. 

y (8)Showthat (4x + 3«/ + l) dx + {3x + 2y + l) dy=0 
represents a family of hyperbolas having as asymptotes the lines 
x + y = and 2x + y + l=0. 

(9) If J + 2«/tanz = sina: 

and y = when x = \ir, show that the maximum value of y is ^. 

[Math. Tripos.] 

(10) Show that the solution of the general homogeneous equation 
of the first order and degree £ =f ( - ) is 

. f dv 

lo g x= \-TT-\ +C » 

6 Jf(v)-V 

where v = y/x. 

(II) Prove that x h y k is an integrating factor of 

py dx + qxdy + x m y n (ry dx + sxdy)=0 

h + 1 Jc + l , h + m + 1 k + n + l 

if = — — and = • 

p q r s 

Use this method to solve 

3y dx -2xdy + x*y- l (\0y dx - 6x dy) = 0. 

(12) By differentiating the equation 

C f(xy) + F(xy) d(xy) +1 *, 
if(xy)-F(xy) xy *y 

Verifythat xy{f(xy)-F(xy)} 



MISCELLANEOUS EXAMPLES 23 

is an integrating factor of 

f(xy) ydx + F (xy) xdy = 0. 
Hence solve (x 2 y 2 + xy + l)ydx-(x 2 y 2 -xy + l)xdy=0. 

(13) Prove that if the equation M dx + N <fo/ = is exact, 

dN = dM 
dx By ' 
[For a proof of the converse see Appendix A.] 

(14) Verify that the condition for an exact equation is satisfied by 

(Pdx + Qdy)e$ Ax)dx = 

Hence show that an integrating factor can always be found for 
Pdx + Qdy = 

if if^.m 

Qldy dx] 
is a function of x only. 
Solve by this method 

(x z + xy 4 ) dx + 2y s dy = 0. 

(15) Find the curve (i) whose polar subtangent is constant ; 

(ii) whose polar subnormal is constant. 

(16) Find the curve which passes through the origin and is such 
that the area included between the curve, the ordinate, and the axis 
of x is k times the cube of that ordinate. 

(17) The normal PG to a curve meets the axis of x in 0. If the 
distance of from the origin is twice the abscissa of P, prove that the 
curve is a rectangular hyperbola. 

(18) Find the curve which is such that the portion of the axis of x 
cut off between the origin and the tangent at any point is proportional 
to the ordinate of that point. 

(19) Find the orthogonal trajectories of the following families of 
curves: (i) (x-l) 2 + y 2 + 2ax = 0, 

•(ii) r = a0, 
(iii) r = a + cos n$, 
and interpret the first result geometrically. 

(20) Obtain the differential equation of the system of confocal conies 
x 2 i y 2 _ 



a 2 + \ b 2 + \ 
and hence show that the system is its own orthogonal trajectory. 

(21) Find the family of curves cutting the family of parabolas 
y 2 = iax at 45°. 



24 DIFFERENTIAL EQUATIONS 

(22) If u + iv =f(x + iy), where u, v, x and y are all real, prove that 
the families u = constant, v = constant are orthogonal trajectories. 

., ., • d 2 u d 2 u x d 2 v d 2 v 

Also prove that 3—0+2-2 = = 5-^ + aTi- 

r ox 1 ay i ox 1 oy* 

[This theorem is of great use in obtaining lines of force and lines of 
constant potential in Electrostatics or stream lines in Hydrodynamics. 
u and v are called Conjugate Functions.] 

(23) The rate of cteca'y~oi radium Is proportional to the amount 
remaining. Prove that the amount at any time t is given by 

A=A Q e~ kt . 

(24) If j =g(} -p) and v = if * = 0, prove that 

v = &tanh %-• 
k 

[This gives the velocity of a falling body in air. taking the resistance 
of the air as proportional to v 2 . As t increases, v approaches the limiting 
value k. A similar equation gives the ionisation of a gas after being 
subjected to an ionising influence for time t. ] 

(25) Two liquids are boiling in a vessel. It is found that the ratio 
of the quantities of each passing off as vapour at any instant is pro- 
portional to the ratio of the quantities still in the liquid state. Prove 
that these quantities (say x and y) are connected by a relation of the 
form y = cx k . 

[From Partington's Higher Mathematics for Students of Chemistry, 
p. 220.] 



CHAPTER III 
LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 

23. The equations to be discussed in this chapter are of the form 

d n y d n ~ x y dy ., . ... 

where f(x) is a function of x, but the p's are all constant. 

These equations are most important in the study of vibrations 
of all kinds, mechanical, acoustical, and electrical. This will be 
illustrated by the miscellaneous examples at the end of the chapter. 
The methods to be given below are chiefly due to Euler and 
D'Alembert.* 

We shall also discuss systems of simultaneous equations of this 
form, and equations reducible to this form by a simple transformation. 

24. The simplest case ; equations of the first order. If we take 
n = l and /(#)=(), equation (1) becomes 

jPo|+Ay=o, (2) 

i.e. p ^+Pidx=0, 

or p log y + p x x = constant, 

so log y = ~Pix/p Q + constant 

- -p^/po+logA, say, 
giving y = Ae~ PlX/Po . 

25. Equations of the second order. If we take n = 2 and f(x) = 0, 
equation (1) becomes 

ftS +A i +fty "° (3) 

* Jean-le-Rond D'Alembert of Paris (1717-1783) is best known by " D'Alem- 
bert' s Principle " in Dynamics. The application of this principle to the motion 
of fluids led him to partial differential equations. 

25 



26 DIFFERENTIAL EQUATIONS 

The solution of equation (2) suggests that y = Ae mx , where m is 
some constant, may satisfy (3). 

With this value of y, equation (3) reduces to 

Ae mx (p Q m 2 +pjm +p 2 ) =0. 
Thus, if m is a root of 

p m 2 +p 1 m+p 2 =0, (4) 

■y = Ae mx is a solution of equation (3), whatever the value of A. 

Let the roots of equation (4) be a and {3. Then, if a and /3 are 
unequal, we have two solutions of equation (3), namely 
y=Ae oX and y=Be px . 
Now, if we substitute y =Ae aZ + Be px in equation (3), we shall get 
Ae aX (p a 2 +p ia +p 2 ) +Be> 3x (p /3 2 +p ± p +p 2 ) =0, 
which is obviously true as a and (3 are the roots of equation (4). 

Thus the sum of two solutions gives a third solution (this might 
have been seen at once from the fact that equation (3) was linear). 
As this third solution contains two arbitrary constants, equal in 
number to the order of the equation, we shall regard it as the general 
solution. 

Equation (4) is known as the " auxiliary equation." 

Example. 

To solve 2 -r\ + 5 —■ + 2y =0 put y = Ae mx as a trial solution. This 

leads to Ae mx (2m 2 + 5m + 2)--=0, 

which is satisfied by m = - 2 or - f . 
The general solution is therefore 

y = Ae- 2x + Be~ ix . 
26. Modification when the auxiliary equation has imaginary or 
complex roots. When the auxiliary equation (4) has roots of the 
form p + iq, p-iq, where i 2 = - 1, it is best to modify the solution 

y=Ae {p+iq)z +Be (p - i ' i)x ) (5) 

so as to present it without imaginary quantities. 

To do this we use the theorems (given in any book on Analytical 
Trigonometry) e nx = cos qx + 1 sm qX} 

e - l i x = cos qx - % sin qx. 
Equation (5) becomes 

y - e px {A (cos qx + i sin qx) +B(cosqx-i sin qx) } 
= e px {E cos qx+F sin qx\ 
writing E for A +B and F for i(A -B). E and F are arbitrary 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 27 

constants, just as A and B are. It looks at first sight as if F must 
be imaginary, but this is not necessarily so. Thus, if 
A = l+2i and B = \ -2i, E = 2 and F = -4. 

Example. ^_ 6 ^ +1%m0 

leads to the auxiliary equation 

m 2 - 6m + 13 = 0, 
whose roots are m = 3 ± 2i. 

The solution may be written as 

• y = Ae^ + ^ x + Be^~ 2i '> x , 
or in the preferable form 

y = e Sx {E cos 2x + F sin 2x), 
or again as ?/ = Ce Zx cos (2# - a), 

' where C cos a = E and C sin a = F, 

so that C = J(E 2 + F 2 ) and tan a = F/E. 

27. Peculiarity of the case of equal roots. When the auxiliary 
equation has equal roots a=/3, the solution 

y = Ae aX +Be fix 
reduces to y = {A + B) e aX . 

Now A + B, the sum of two arbitrary constants, is really only a 
single arbitrary constant. Thus the solution cannot be regarded as 
the most general one. 

We shall prove later (Art. 34) that the general solution is 
y = (A+Bx)e aX . 

28. Extension to orders higher than the second. The methods 
of Arts. 25 and 26 apply to equation (1) whatever the value of n, as 
long as/(x)=0. 

The auxiliary equation is 

m 3 -6m 2 + llm-6 = 0, 
giving m = l, 2, or 3. 

Thus y = Ae x + Be 2x + Ce 3x . 

Ex.(ii). U' 8y = °' 

The auxiliary equation is m 3 - 8 = 0, 

i.e. (m-2)(m 2 + 2m + 4)=0, 
giving m = 2 or -l±i\/S. 

Thus y = Ae 2x + e~ x (E cos x^/3 + F sin x\/3), 

or y = Ae 2x + Ce~ x cos (x\/3- a). 



28 DIFFERENTIAL EQUATIONS 

Examples for solution. 

Solve 



y,., d 2 s .ds J d 2 s . ds _ 

(7) «^ + 2 &i & 2y_a 

(8) What does the solution to the last example become if the initial 
conditions are fly 

y = l, -p = when x = 0, 

and if y is to remain ^finite when x= + co ? 
Solve 

<»»3-»3 + "»-*- v t . 

•(H) ^ + 8y=0. \| .(12)g-64<,=0. 

,72/3 J/3 

* (13) Z-T-g +#0=0, -given that = a and tt=0 when t=0. 

[The approximate equation for small oscillations of a simple pen- 
dulum of length I, starting from rest in a position inclined at a to the 
vertical.] 

(14) Find the condition that trigonometrical terms should appear 
in the solution of ^2 S fe 

m dT* +k di +cs=0 - 

[The equation of motion of a particle of mass m, attracted to a 
fixed point in its line of motion by a force of c times its di -ance from 
that point, and damped by a frictional resistance of k times its velocity. 
The condition required expresses that the motion should be oscillatory. 
e.g. a tuning fork vibrating in air where the elastic force tending to 
restore it to the equilibrium position is proportional to the displacement 
and the resistance of the air is proportional to the velocity.] 

(15) Prove that if k is so small that k 2 /mc is negligible, the solution 
of the equation of Ex. (14) is approximately e~ kt/2m times what it would 
be if k were zero. 

[This shows that slight damping leaves the frequency practically 
unaltered, but causes the amplitude of successive vibrations to diminish 
in a geometric progression. ] 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 29 

(16) Solve L^ + R^ + Q-O, given that Q=Q Q and ^ = when 

* = 0, and that CR 2 <4:L. 

[Q is the charge at time t on one of the coatings of a Leyden jar of 
capacity C, whose coatings are connected when t = by a wire of resist- 
ance R and coefficient of self-induction L. ] 

29. The Complementary Function and the Particular Integral. So 
far we have dealt only with examples where the f(x) of equation (1) 
has been equal to zero. We shall now show the relation between 
the solution of the equation when f(x) is not zero and the solution 
of the simpler equation derived from it by replacing f(x) by zero. 
To start with a simple example, consider the equation 

It is obvious that y=x is one solution. Such a solution, con- 
taining no arbitrary constants, is called a Particular Integral. 
Now if we write y=x+v, the differential equation becomes 

2g + R(l. + £) + «t, + .>-5 + * 

n d%* K dv _ n 

' giving v = Ae~ 2x + Be~ ix , 

T so that y=x+Ae- 2x +Be-*- x . 

The terms containing the arbitrary constants are called the 
Complementary Function. 

This can easily be generalised. 
If y = u is a particular integral of 

d n y {? n_1 v dy £ . » ia . 

*lf+AjSA + ~ + **£ + **-fW (6) 

so that Po^a+Pi^=i + -+Pn-i-^.+PnU=f(x), (7) 

put y=u + v in equation (6) and subtract equation (7). This gives 
d n v d n ~ x v dv _ /ox 

V°fan+Pl^+---+Pn-l lx +PnV=0 (8) 

If the solution of (8) be v = F(x), containing n arbitrary con- 
stants, the general solution of (6) is 

y = u+F(x), 
and F(x) is called the Complementary Function. 



30 DIFFERENTIAL EQUATIONS 

Thus the general solution of a linear differential equation with 
constant coefficients is the sum of a Particular Integral and the Com- 
plementary Function, the latter being the solution of the equation 
obtained by substituting zero for the function of x occurring. 

Examples for solution. 

Verify that the given functions are particular integrals of the follow- 
ing equations, and find the general solutions : 

j I *•> p-4 + *>-<"- < 2 > 3 ■• - g- i3 ! +i » 

• (3)2sin3a;; j\ + iy = - 10 sin Sx. 

For what values of the constants are the given functions particular 
integrals of the following equations ? ^ 

(4)ae»*; g + 13^ + 42y»ll2e» ^,>V' X 

d 2 s \j / d 2 v 

- ( 5 ) aeU > ij2 +9s= QOe ~ t - &v ( 6 ) a sin px ' ri + y = 12 sin 2cc * 

V-^7) a sin px + b cos px ; -=-| + 4 j^ + 3y = 8 cos x - 6 sin x. 

<«> •' § +5 l +6 ^ 12 - 

Obtain, by trial, particular integrals of the following : 

.(11) + 9y = 4Osin5*. < 12 > H" 8 l + 9 !' = 40sill5a; - 

» < 13) § + 8 ! + 2 » 

30. The operator D and the fundamental laws of algebra. When 
a particular integral is not obvious by inspection, it is convenient 
to employ certain methods involving the operator D, which stands 

for -j-. This operator is also useful in establishing the form of the 

complementary function when the auxiliary equation has equal 

roots. 

d 2 d 3 

D 2 will be used for j- 2 , D 3 for -7-3, and so on. 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 31 

The expression 2 -y-| + 5 -j- + 2y may then be written 

2D*y + 5Dy+2y, 
or (2D*+bD + 2)y. 

We shall even write this in the factorised form 
(2D + l)(D+2)y, 
factorising the expression in D as if it were an ordinary algebraic 
quantity. Is this justifiable ? 

The operations performed in ordinary algebra are based upon 
three laws : 

I. The Distributive Law 

m(a + b)=ma + mb ; 
II. The Commutative Law 

ab = ba ; 
III. The Index Law a m . a n =a m+n . 
Now D satisfies the first and third of these laws, for 
D(u+v)=Du+Dv, 
and D m . D n u=D m+n . u 

(m and n positive integers). 
As for the second law, D (cu) = c (Du) is true if c is a constant, 
but not if c is a variable. 

Also D m (D n u) = D n (D m u) 

(m and n positive integers). 
Thus D satisfies the fundamental laws of algebra except in that 
it is not commutative with variables. In what follows we shall 
write F(D) s p D» + Pl D^ + ... +p n _ 1 D +p n , 

where the p's are constants and n is a positive integer. We are 
justified in factorising this or performing any other operations 
depending on the fundamental laws of algebra. For an example 
of how the commutative law for operators ceases to hold when 
negative powers of D occur, see Ex. (iii) of Art. 37. 

31. F(D)e ax =e ax F(a). Since 

De ax = ae ax , 

D 2 e ax =a 2 e ax , ' 

and so on, ,^ 

F(D) e ax = (p D n +p 1 D n ~ 1 + ... +p n - x I) +p n ) e ax 

^(p^+p^- 1 + ... +p n . 1 a +p n ) e ax 

=e a *F(a).\ 



32 DIFFERENTIAL EQUATIONS 

32. F(D){e ax V} =e ax F(D +a) V, where V is any function of x. By 
Leibniz's theorem for the n th differential coefficient of a product, 
D»{eP*V} = (D n e ax ) V + w(D"- 1 e oa ') {DV) 

+ ln(n-l)(D n -*e ax )(D*V) + ... + e ax (D n V) 
=a n e°*Y +na n - 1 e az DV +\n{n - l)a»-V*D a 7 + ... +e ax D n V 
=e ax (a n +na n r 1 I>k + hn(n - l)a n ~ 2 D 2 + ...+D n )V 
= e a *(Z)+a) n F. 
Similarly D n ~ 1 {e ax V}=e ax (D+a) n - 1 V, and so on. 
Therefore 
F(D){e ax V} = (p D» + Pl D»~ l + ... +p n . 1 D +p n ){e ax V} 

= e ax {p (D+a) n +p 1 (D+a) n ~ 1 + ... +p n - x {D +a) +p n }V 
= e* x F(D + a)V. 

33. F(D 2 ) cos ax =F( - a 2 ) cos ax. Since 

D 2 cos ax = -a 2 cos ax, 

D i oosax = (- a 2 ) 2 cos ax, 
and so on, 

F(D 2 ) cos ax = (p D 2n +p 1 D 2n ~ 2 + ... +p n - 1 D 2 +p n ) cos ax 

= {Po(~ a2 ) n +Pi(-a 2 ) n ~ 1 + ... +p n -i( -« 2 ) +Pn} cos «# 
— F( -a 2 ) cos ax. 
Similarly F (D 2 ) sinax=F(- a 2 ) sin ax. 

34. Complementary Function when the auxiliary equation has equal 
roots. When the auxiliary equation has equal roots a and a, it 
may be written m 2 _ 2 ma + a 2 = 0. 

The original differential equation will then be 

i.e. (D 2 -2aZ>+a 2 )*/=0, 

(D-a) 2 y=0 :...(9) 

We have already found that y=Ae aX is one solution. To find 
a more general one put y=e*?V, where V is a function of x. 
By Art. 32, 

(D -a)*{e?*V} =e aX (D -a +a) 2 V = (T?D 2 V. 
Thus equation (9) becomes 

D 2 V=0, 
i.e. V = A+Bx, 
so that y = e ax (A+Bx). 



m 



LINEAR EQL\. X)NSTANT COEFFICIENTS 33 

Similarly the equation (D - a) p y = 
reduces to D P V =0, 

giving V = {A x + A& + A &? + . . . + Ap^' 1 ), 

and y = e aX (A x + A^c + A z x 2 + . . . + A p ^xP^ x ). 

When there are several repeated roots, as in 

(D-anD-(3)*(D-yy y =0, (10) 

we note that as the operators are commutative we may rewrite the 
equation in the form 

iD-PnD-yY{{D-aVy}=0, 
which is therefore satisfied by any solution of the simpler equation 

(D- a y y =o '. (ii) 

Similarly equation (10) is satisfied by any solution of 

(D-P)*y=0, (12) 

or of (D-y) r y=0 (13) 

The general solution of (10) is the sum of the general solutions 
of (11), (12), and (13), containing together (p+q+r) arbitrary 
constants. 



Ex. (i). Solve (D*-8D 2 + 16)y = 0, (p. 



^ 



i.e. (Z) 2 -4) 2 ?/ = 0. 
The auxiliary equation is (m 2 -4) 2 = 0, 

m = 2 (twice) or -2 (twice). 
Thus by the rule the solution is 

y = (A + Bx) e 2x + (E + Fx) e~ 2x . 

Ex. (ii). Solve (D 2 + l) 2 */ = 0. 

The auxiliary equation is (m 2 + l) 2 = 0, 

m — i (twice) or -i (twice). 
Thus y = {A + Bx)e ix + {E+Fx)e~ ix , 

or better y = (P+ Qx) cos x + ( R + Sx) sin x. 

Examples for solution. 

\h) {D* + 2D* + D 2 )y = 0. >< $) (-D 6 + 3Z>* + 32) a + l)y = 0. 

^) (D i -2D 3 + 2D 2 -2D + l)y = 0. $4) (4Z> 5 -3Z> 3 - D 2 ) t/ = 0. s 

(5) Show that 

F (D 2 ){P cosh ax + Q sinh ax) = F(a 2 ) (P cosh ax + Q sinh ax) . 

(6) Show that (D - a) 4n (e ax sin px) =p* n e ax sin px. 

35. Symbolical methods of finding the Particular Integral when 
f(x) =e ax . The following methods are a development of the idea 
of treating the operator D as if it were an ordinary algebraic quan- 

p.d.e. c 



34 DIFE 

tity. We shall proceed tentatively, at first performing any opera- 
tions that seem plausible, and then, when a result has been obtained 
in this manner, verifying it by direct differentiation. We shall use 

the notation Y7Jy\ /(*) *° denote a particular integral of the equati 
„ F(D)y-f{z). 



(i) If f(x) =e ax , the result of Art. 31, 

F(D) e™ = e ax F (a) 

1 



" 



suggests that, as long as F(a)=j=0, ^y-r e ax may be a value of 



F(u) v — v_—"» F ( D y 

This suggestion is easily verified, for 



W^}-9lSrVyArt.3i. 



>F(a) J F{a) 
= e ax . 
(ii) If F (a) =0, {D-a) must be a factor of F(D). 
Suppose that F(D)=(D -a) p <f>(D), where 0(a)=/=O. 
Then the result of Art. 32, 

F (D) {e ax V } = e ax F (D + a) V, 
suggests that the following may be true, if 7 is 1, 

1-^^ 1 _, 1 l(* x .l\ e™ 1 , 

/jw*~--. pax j y _ 

F(D) (D-a)P<f>(D) (D-a)p\</>(a){ <j>(a) Dp 

e? x x v 

adopting the very natural suggestion that jz is the operator inverse 
to D, that is the operator that integrates with respect to x, while 
y- integrates p times. Again the result obtained in this tentative 
manner is easily verified, for 

■♦^K^S' byArt - 32 ' 

= e? x , byArt. 31. 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 35 

In working numerical examples it will not be necessary to repeat 
the verification of our tentative methods. 

Ex. (i). (D + 3) 2 y = 50e* x . 

The particular integral ia 



(Z> + 3) 2 (2+3) 

Adding the complementary function, we get 
y = 2e 2x + {A + Bx)e~ 3x . 
Ex. (ii). (Z>-2) 2 i/=.50e 2 *. 

If we substitute 2 for D in jj: — ^ 50e 2x , we get infinity. 
But using the other method, 

(D ]_ 2)2 • 50e 2 * = 50e 2ie -^- 2 . 1 = 5Qe 2 * , \x 2 - 25x 2 e 2 *. \ 

Adding the complementary function, we get 
y = 25x 2 e 2x + (A + Bx)e 2x . 

Examples for solution. 
Solve v? 

m)(D 2 + 6D + 25)y=^lQi:^ x . rfft) (D 2 + 2pD + p 2 + q 2 )y = e ax . 

:($) (D 2 -9)?/ = 54^^ m (D 3 -D)y = e x + e- x . 

(5) (D 2 -p 2 )y = a'2o&px. (6) (D* + iD 2 + lD) y = 8e~ 2x . 

36. Particular Integral when f (x) =cos ax. From Art. 33, 

(j> (D 2 ) cos ax = <f> ( - a 2 ) cos ax. 
This suggests that we may obtain the particular integral by 
writing - a 2 for D 2 wherever it occurs. 

Ex. (i). (D 2 + SD + 2) y = cos 2x. 

1 1 1 

. COS 2x = ; t-=: - . COS 2x = 7rF : rr . cos 2x. 



D 2 + 3D + 2 -4 + 3Z) + 2 3D-2 

To get D 2 in the denominator, try the effect of writing 

1 _ 3Z> + 2 
3ZT^2~9Z> 2 -4' 

suggested by the usual method of dealing with surds. 
This gives 

— ^r — i cos 2x = - xV(3-D cos 2x + 2 cos 2x) 
- oo - 4 

= - t L .(-6sin2a; + 2cos2a;) 

^^(Ssu^aj-cc^a;). 



36 DIFFERENTIAL EQUATIONS 

Ex. (ii). (D i + 6D 2 + nD + 6)y = 2ein3x. 

2 sin 3x = 2 — ^= — — : — r-r^ — - sin 3x 



D 3 + 6D 2 + llD + 6 -9D-54- LD + 6 

1 



Z)-24 
D + 24 



Z> 2 -576 



_ i 



sin 3z 
sin 3x 



5 s 



-(3 cos 3a; + 24 sin Sx) 



= - T -^ T (cos 3x + 8 sin 3x). 
We may now show, by direct differentiation, that the results 
obtained are correct. 

If this method is applied to 

[<f> (D 2 ) + D\J, (D 2 ) ] y = P cos ax + Q sin ax, 
where P, Q and a are constants, we obtain 

<j> ( - a 2 ) . (P cos ax + Q sin ax)+a\p- {-a 2 ) . (P sin ax -Q cos ax) 
{^(-a 2 )} 2 + a 2 {yjr(-a 2 )} 2 . 

It is quite easy to show that this is really a particular integral, 
provided that the denominator does not vanish. This exceptional case 
is treated later (Art. 38). 

Examples for solution. 

Solve / 

.$& {D + l)y = 10sm2x. fa (Z) 2 -5Z) + 6) </ = 100sin ix. 

^(3) (Z) 2 + 8D + 25)i/ = 48 cos a: -16 sin a. 
v <4) (D 2 + 2D + 401) y = sin 20z + 40 cos 20x. 

(5) Prove that the particular integral of 

d 2 s _ 7 ds „ 

-tt + 2* -t; + V s = a cos at 

at 2 at 

may be written in the form b cos (qt - e), 

where b = a/{(p 2 -q 2 ) 2 + ik 2 q 2 } h and tan € = 2kq/(p 2 -q 2 ). 

Hence prove that if q is a variable and &, p and a constants, 6 is 
greatest when q = y/{p 2 - 2k 2 ) = p approx. if k is very small, and then 
e = 7r/2 approx. and b = a/2kp approx. 

[This differential equation refers to a vibrating system damped 
by a force proportional to the velocity and acted upon by an external 
periodic force. The particular integral gives the forced vibrations 
and the complementary function the free vibrations, which are soon 
damped out (see Ex. 15 following Art. 28). The forced vibrations 
have the greatest amplitude if the period 2-n-fq of the external force 
is very nearly equal to that of the free vibrations (which is 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 37 

2ir/y/{p 2 -k 2 )=2ir/p approx.), and then e the difference in phase 
between the external force and the response is approx. tt/2. This 
is the important phenomenon of Resonance, which has important 
applications to Acoustics, Engineering and Wireless Telegraphy.] 

37. Particular integral when f(x) =x m , where m is a positive integer. 

In this case the tentative method is to expand j^ in a series of 
ascending powers of D. 

Ex.(i). -^^ = 1.(1+^)-^ 

= i(* 2 -i). 
Hence, adding the complementary function, the solution suggested 

for {D*+4)y = z* 

is y = \(x 2 -^)+A cos 2x4-5 sin 2x. 

Ex. (ii). 

D 2 -iD + S ^ = £ \i^d ~ 3TI)) **> b ? P artial frac tions, 

f / D D 2 D 3 Z) 4 M 

= i {(l + Z) + Z) 2 + Z)3 + Z). 4 + ...)- i (l + - + T+ - + - + ...)}^ 

Adding the complementary function, the solution suggested for 
(D 2 -iD + 3)y = x 3 
is y^^xS + ^ + zg-x + ^ + Aet + Be?*. 

s* m dhb^T) 96x2=w ■ Uwrt x °} 

-W.-jTj^-g). from Ex. (i), 

-■i(S-t) 

= 2x 4 -6x 2 . 
Hence the solution of D 2 (Z) 2 + 4) y = 96x 2 should be 

y = 2x*-6x 2 + A cos 2x + Z?sin 2x + E + Fx. 
Alternative method. 

= (24:D- 2 -6 + %D 2 -...)x 2 
= 2x*-6x 2 + 3. 



38 DIFFERENTIAL EQUATIONS 

This gives an extra term 3, which is, however, included in the 
complementary function. 

* The method adopted in Exs. (i) and (ii), where F(D) does not 
contain D as a factor, may be justified as follows. Suppose the expan- 
sions have been obtained by ordinary long division. This is always 
possible, although the use of partial fractions may be more convenient 
in practice. If the division is continued until the quotient contains D m , 
the remainder will have D m+1 as a factor. Call it <p(D) . D m+1 . Then 

^= CQ + c 1 D + c 2 D* + ...+c m n™ + * iD ] }- ] ^ +1 (1) 

This is an algebraical identity, leading to 

l = F(D){c + c 1 D + c t D* + ...+c m D m } + <f>{D) . D"* 1 (2) 

Now equation (2), which is true when D is an algebraical quantity, 
is of the simple form depending only on the elementary laws of algebra, 
which have beffc shown to apply to the operator D, and it does not 
involve the difficulties which arise when division by functions of D is 
concerned. Therefore equation (2) is also true when each side of the 
equation is regarded as an operator. Operating on x m we get, since 
D m+1 x m = 0, 

x m =F(D){{c (t + c 1 D + c 2 D 2 + ...+c m D m )x m }, (3) 

which proves that the expansion obtained in (1), disregarding the 
remainder, supplies a particular integral of F(D)y=x m . 

It is interesting to note that this method holds good even if the 
expansion would be divergent for algebraical values of D. 

To verify the first method in cases like Ex. (iii), we have to prove 
that 1 

i.e. (c D- r + c 1 D~ r + 1 + c 2 D- r + 2 + ... +c m D~ r + m ) x m , 
is a particular integral of {F(D) . D r }y = x m , 

i.e. that {F (D) . D r } {(c Q D- r + c^* 1 + c 2 D~ r+2 

+ ...+c m D- r+m )x m }=x m (4) 

Now {F( D) . D'} u m F(D) . {D r u}, 

also D r {(c 8 D ~ '■+*) x m } = (c g D*) x m ; 

hence the expression on the left-hand side of (4) becomes 

F(D){(c + c 1 D + c 2 D* + ...+c m D™)x m } = x'», by (3), 

which is what was to be proved. 

In the alternative method we get r extra terms in the particular 
integral, say ( Cffi+1 Z)^^+...+ W 2>»)*« 

These give terms involving the (r-l) th and lower powers of x. 
But these all occur in the complementary function. Hence the first 
method is preferable. 

* The rest of this article should be omitted on a first reading. 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 39 

Note that if D~hi denotes the simplest form of the integral of u, 
without any arbitrary constant, 

D- 1 (DA) = D- 1 . 0=0, 

while i>{2H.l)-X>.*-l, 

so that D(D- 1 .l)^D- 1 .(Z).l). 

Similarly D m (D- m . x n )=^D- m (D m . x n ), if m is greater than n. 

So when negative powers of D are concerned, the laws of algebra 
are not always obeyed. This explains why the two different methods 
adopted in Ex. (iii) give different results. 

Examples for solution. 

Solve 
-j(l) (D + \)y = %*. /(2) (D 2 + 2D)y = 2ix. 

^(3) (D 2 -6D + 9)y = 5±x + l8. ' (4) (D 4 -6D 3 + 95 2 )?/ = 54a: + 18. , 

v/(5) (D 2 -D-2)y = U-76x-±8x 2 . 

JG) (IP-D 2 -2D)y = U-76x-±8x 2 . U 

38. Particular integrals in other simple cases. We shall now 
give some typical examples of the evaluation of particular integrals 
in simple cases which have not been dealt with in the preceding 
articles. The work is tentative, as before. For the sake of brevity, 
the verification is omitted, as it is very similar to the verifications 
already given. 

Ex. (i). (D 2 + 4)y=sm2x. 

We cannot evaluate -^ — j sin 2x by writing - 2 2 for D 2 , as in 

Art. 36, for this gives zero in the denominator. 
But i sin 2x is the imaginary part of e 2ix , and 

e ax = e 2ix t i f as m Art. 35, 



Z) 2 + 4 (Z> + 2*') 2 + 4 



— ■ x _ JL 



L_.i ±-S*> -x.i 



1 I. Dv* 



c a ". 



-'"OT-{(»-B+™--)-U P) 



" e 4*D 4* 

«■ - |«x(cos 2a; + 1 sin 2x) ; 



40 DIFFERENTIAL EQUATIONS 

hence, picking out the imaginary part, 

™ — r sin 2x = - \x cos 2x. 

Adding the complementary function, we get 

y = A cos 2x + B sin 2x - \x cos 2x. 
Ex. (ii). (D 2 - 52) + 6) y = e 2x x*. 

(D2-5D + 6) \2-D 3-D/ ^ 

= e2x (4-r^)^ * 



= e 2 * ( - ^ - 1 - D - D 2 - D 3 - D* - ...) 



a? 



= e 2 *( - \x* - x 3 - 3x 2 - 6x - 6). 
Adding the complementary function, we get 

y = A<? x - e**{\x* + x 3 + 3x* + 6x- B), 
including the term - 6e 2x in Be 2x . 

Ex. (iii). (Z) 2 - 6D + 13) y = 8<? x sin 2x. 

. 8<? x sin 2x = 8e Sx ,,„ ..,„ * -^ — r^ . sin 2x 



(D 2 -6D + 13)" {(D + 3) 2 -6(Z) + 3) + 13} 

= 8e 3a: -K5 — 7sin2z 

= 8e 3a: ( - \x cos 2x) (see Ex. (i) ) 
= - 2xe? x cos 2x. 
Adding the complementary function, we get 

y = e? x (A cos 2x + B sin 2x - 2x cos 2x). 

These methods are sufficient to evaluate nearly all the particular 
integrals that the student is likely to meet. All other cases may 
be dealt with on the lines indicated in (33) and (34) of the miscel- 
laneous examples at the end of this chapter. 

Examples for solution. 

Solve 
"(1) (Z) 2 + l)</ = 4cosa;. (2) (D-l) y = (x + 3) e* x . 

y\Z) (D !i -3 L D-2)y = 5i0x 3 e- x . y(i) ( DM- 2D + 2) y » 2e-*«n x. f : 

(5) {D^+}J^ji=2ixcosx. (6) (W^~D)y = \2e x + 8 sin x -2 x. 

(7) (D 2 -6D + 25)y = 2e 3 *cos4:r + 8e 3a: (l-2:r)sin4x. . 

39. The Homogeneous Linear Equation. This is the name given 
to the form {p ( fc n B n +p 1 x n ~ 1 D n - 1 + . . . + p n ) y =f (x). 

It reduces to the type considered before if we put x = e'. 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 41 

Ex. (x 3 D 3 + 3x 2 D 2 + xD)y = 2ix 2 . 

Put x = e l , 

dx_ t _ 
dt~ 6=X> 

t . _. d dt d 1 d 

so that D = dZ=" d - x jr-J t ' 

\xdt) x*dt + x dt x 2 \ dt^dt 2 ).' 
x 2 \ dt + dt 2 ) a*\ dt + dt 2 J + x 2 \ dt + dt 2 / 
x*\ dt + dt 2 ) + x 3 \ dt 2 + dt?) 
x*\ dt dt 2 dtV' 

d u 

thus the given differential equation reduces to -^ = 2ie 2 \ 

giving y = A + Bt + Ct 2 + 3e it 

= A + B\ogx + C(logx) 2 + 3x 2 . 

Another method is indicated in (28)-(30) of the miscellaneous 
examples at the end of this chapter. 
The equation 

p (a+bx) n D n y +p 1 (a + bx) n - 1 D n ~ 1 y + ... +p n y =f(x) 
can be reduced to the homogeneous linear form by putting 
z = a +bx, giving ^ = dy^dydz =b dy 

y dx dz dx dz ' 

Examples for solution. 
/(I) x 2 ^-2x d £ + 2y = M. / (2) ^g + 9x| + 25^50. 

(3) x3 ^ + 3:K2 § + a; | + 8 ^ = 65c0s(l0 ^ ) - 

W ^dx* + Z ^dx* + X dx 2 X dx + y ~ [ ° gX - 

(5) (1+2^2-6(1 +2x)g + 16^ r 8 (l+2a:)«. 

(6) (l + x) 2< ^ + (l+x) d £ + y==ico8log(l+x). 



42 DIFFERENTIAL EQUATIONS 

40. Simultaneous linear equations with constant coefficients. The 
method will be illustrated by an example. We have two de- 
pendent variables, y and z, and one independent variable x. 

D stands for -»-, as before. 
ax 

Consider (52)+4) y -(2D + l)z = e~ x , (1) 

(D + S)y- 3z =be~ x (2) 

Eliminate z, as in simultaneous linear equations of elementary 
algebra. To do this we multiply equation (1) by 3 and operate on 
-equation (2) by (22) + 1). 

Subtracting the results, we get 

{3 (52) +4) - (22) + 1)(D + 8)} y = Se~ x - (2D + 1) 5e~*, 
i.e. (-2D 2 -2D + ±)y = Ser* t 
or (D 2 jD-2)y = -4<r* 

Solving this in the usual way, we get 

y = 2e- x +Ae x +Be~ 2x . 
The easiest way to get z in this particular example is to use 
•equation (2), which does not involve any differential coefficients of z. 
Substituting for y in (2), we get 

Ue~ x + 9Ae x + QBe~ 2x - 3z = 5e~*, 
so that z = 36-* + 3Ae x + 2Be~ 2x . 

However, when the equations do not permit of such a simple 
method of finding z, we may eliminate y. 
In our case this gives 
{ - (2) +8) (2D + 1) + 3(52) + 4)} y = {D + S)e~ x - (52) + 4)5e"*, 
i.e. (-22) 2 -22) + 4)z = 12e-*, 
-giving z = 3e~ x + Ee x + Fe~ 2x . 

To find the relation between the four constants A, B, E, and F, 
substitute in either of the original equations, say (2). This gives 
(2) + 8) (2e~ x + Ae x + Be~ 2x ) - 3 {3e~ x + Ee x + Fe~ 2x ) = 5e~ x , 
i.e. (9 A - ZE) e x + (6# - 32?) e~ 2x = 0, 
whence E = ?>A and F = 2B, 

so z = Se~ x + Ee x + Fe~ 2x = 3e~ x + 3^e* + 2Be~ 2x , as before. 

Examples for solution. 

(1) Dy-z = 0, (2) (Z>- 17) t/ + (22)- 8)2 = 0, 

(D-l)y-(D + \)z = 0. , (132)- 53)?/- 22 = 0. 

<3) (22) 2 -2) + 9)y-(Z) 2 + 2) + 3)2 = ) 
(22) 2 + D + 7) y -{D 2 -D + 5)2=0. 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 43 

<4) (D + \)y = z + e x , (5) {D 2 + b)y-d = -36cos 7x, 

{D + l)z = y + e x . p>n) 2 z = 99co8 7x. 

(6) (2.D + l)«/ + (Z> + 32)z = 91e-* + 147sin2a; + 135cos2a;, 
y-(D-8)z = 29e~ x + 47 sin 2x + 23 cos 2x. 




MISCELLANEOUS EXAMPLES ON CHAPTER III. 

Solve 
/O) (Z)-l) 3 2/ = 16e 3 *. (2) (4D 2 + 12Z) + 9)i/ = 144a*fH 

u <3) (D* + 6IP + nD 2 + 6D) y = 20e~ 2x sin x. 

<4) (D 3 -D 2 + 4:D-i)y = 68e x am2x. 

(5) (D*-6D 2 -8D-S)y = 256(x + l)(? x . 

(6) (ZH-8Z) 2 -9)i/ = 50sinh2a;. (7) (Z>»-2Z) 2 + 1) */ = 40cosh x. 
(8) (Z)-2) 2 2/ = 8(x 2 + e 2 * + sin2x). (9) (D-2) 2 y = 8x 2 e 2x Bin 2x. 

<10) (7) 2 + l)?/ = 3cos 2 a; + 2 8in 3 x. 

<11) (D* + lOD 2 + 9)y = 96am2xcosx. 

(12) (D-a) a y = a x , where a is a positive integer. 

ax 2 x ax x 2 dx 2 x dx 

< 15 > % = f ^ (, + l) 2 g + (, + l)gH2x + 3)(2x + 4). 

§ + 4§ + 4, = 25* + W. 
«**-* £-*;>* <l«)£ + I-0; <§ + *-0. 

(21) Show that the solution of (D 2n + 1 -l)y = consists of Je* and 
n pairs of terms of the form 

ef x (B r cos sx + C r sin sx), 

. 2ttt . . 2xr 

where c = cos ^ and s = sin - , 

2w + 1 2n + 1 

r taking the values 1, 2, 3 ... n successively. 

(22) If (D-a)u = 0, 

(D-a) v = u, 
and (D- a)y = v, 

find successively w, w, and y, and hence solve (D-a) 3 </ = 



/ 



44 DIFFERENTIAL EQUATIONS 

(23) Show that the solution of 

(D-a)(D-a-h)(D-a-2h)y=0 

can be written Ae ax + B^ x - — r — -+Ce ax - ^ '-. 

h h* 

Hence deduce the solution of (D- a) z y = 0. 

[This method is due to D'Alembert. The advanced student will 
notice that it is not quite satisfactory without further discussion. It 
is obvious that the second differential equation is the limit of the first, 
but it is not obvious that the solution of the second is the limit of the 
solution of the first.] 

oz o z 

(24) If {D-a) 3 e mx is denoted by z, prove that z, =— , and =— ^ all 

vanish when m = a. 

Hence prove that e ax , xe? x , and x 2 e ax are all solutions of (D - a) 3 y =0. 

[Note that the operators {D-af and -~— are commutative.] 

.-„, .., , , cos ax - cos (a + h) x 

(25 Show that ; ,. 2 v . 

x (a + h) 2 -a 2 

is a solution of (D 2 + a 2 ) y = cos (a + h) x. 

Hence deduce the Particular Integral of (D 2 + a 2 )y = cos ax. 

[This is open to the same objection as Example 23.] 

(26) Prove that if V is a function of x and F(D) has its usual 
meaning, 

(i) D n [xV] =xD n V + nD n -W\ 
(ii) F(D)[xV] = xF{D)V + F'(D)V; 

(iii)^-rsri-s — 7- F{D) v- 

{m} F(D) [ - J F(D) V [F(D)] 2 ' 

(iv) <p(D)[x n V] = x n tp{D)V + nx n - 1 </>'(D)V + --- + n C,.x n - r <f> r (D)V 
where <p{D) stands for • 

(27) Obtain the Particular Integrals of (i) (D- l)y = xe 2x , 

(ii) (D + l)y = x 2 coax, 
by using the results (iii) and (iv) of the last example. 

(28) Prove, by induction or otherwise, that if 6 stands for x 3-, 

x»^=e(d-l)(0-2)...(0-n + l)y. 

(29) Prove that 

(i) F(6)x m = x m F{m); 
x m 

(u) jrgf ~r$sj- p rovided F ( m )±°'- 

(iii » m [xmV]= ^-FW^) v - 

where V is a function of x. 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 45 

(30) By using the results of the last question, prove that the solu- 
tion of d2 d ..-,-. „ „ 

x 2 j\-ix~+%y = x 5 is $x 5 + Ax a + Bx b , 

-where a and b are the roots of m{m - 1) - 4m + 6 = 0, 

i.e. 2 and 3. 

(31) Given that (D-l)y = e 2x , 

prove that (D-l)(D-2)y = 0. 

By writing down the general solution of the second differential 
equation (involving two unknown constants) and substituting in the 
first, obtain the value of one of these constants, hence obtaining the 
solution of the first equation. 

d 2 y 

(32) Solve j^ 2 +p 2 y = sin ax by the method of the last question. 

(33) If u x denotes e ax I ue~ ax dx, 

u 2 denotes e bx I u^er** dx, 

etc., 
prove the solution of F(D)y = u, where F(D) is the product of n 
factors. 

(D-a)(D-b)... 

may be written y = u n- 

This is true even if the factors of F(D) are not all different. 
Hence solve (D-a)(D-b)y = e ax log x. 

(34) By putting „ ~ into partial fractions, prove the solution of 
F(D)y = u may be expressed in the form 

2-^rrre aa; l ue~ ax dx, 
F'{a) J 

provided the factors of F(D) are all different. 

[If the factors of F(D) are not all different, we get repeated inte- 
grations.] 

Theoretically the methods of this example and the last enable us to 
solve any linear equation with constant coefficients. Unfortunately, 
unless u is one of the simple functions (products of exponentials, sines 
and cosines, and polynomials) discussed in the text, we are generally 
left with an indefinite integration which cannot be performed. 

If u =f(x), we can rewrite e ax I ue~ ax dx 

in the form f(t)e a ^-^dt, 

where the lower limit Jc is an arbitrary constant. 



46 DIFFERENTIAL EQUATIONS 

(35) (i) Verify that 

1 f * 
y = ~\ f(t)ainp(x-t)dt 

is a Particular Integral of 

[Remember that if a and b are functions of x, 

(ii) Obtain this Particular Integral by using the result of the last 
example. 

(iii) Hence solve (Z) 2 + l)i/ = cosec x. 

(iv) Show that this method will also give the solution of 

(in a form free from signs of integration), if f(x) is any one of the func- 
tions tan x, cot x, sec x). 

(36) Show that the Particular Integral of j~ + p 2 y = k cos pt repre- 
sents an oscillation with an indefinitely increasing amplitude. 

[This is the phenomenon of Resonance, which we have mentioned 
before (see Ex. 5 following Art. 36). Of course the physical equatic 
of this type are only approximate, so it must not be assumed thatjpie 
oscillation really becomes infinite. Still it may become too^pge 
for safety. It is for this reason that soldiers break step on c^£hg a 
bridge, in case their steps might be in tune with the natural osOTlation 
of the structure. ] 

(37) Show that the Particular Integral of 

-^ + 2h-^ + (h 2 + p 2 )y = Jce- ht cospi 

k 
represents an oscillation with a variable amplitude —te~ ht . 

Find the maximum value of this amplitude, and show that it is very 
large if h is very small. What is the value of the amplitude after an 
infinite time ? 

[This represents the forced vibration of a system which is in reson- 
ance with the forcing agency, when both are damped by friction. The 
result shows that if this friction is small the forced vibrations soon 
become large, though not infinite as in the last example. This is an 
advantage in some cases. If the receiving instruments of wireless 
telegraphy were not in resonance with the Hertzian waves, the effects 
would be too faint to be detected.] 



LINEAR EQUATIONS WITH CONSTANT COEFFICIENTS 47 

(38) Solve ^~ n * y=0 - 

[This equation gives the lateral displacement y of any portion of a 
thin vertical shaft in rapid rotation, x being the vertical height of the 
portion considered. ] 

(39) If, in the last example, 

-i = y=0 when x=0 and x = l, 

prove that y = E(coa nx - cosh nx) + F(ain nx - sinh nx) 
and cos nl cosh nl = 1 . 

[This means that the shaft is supported at two points, one a height 
I above the other, and is compelled to be vertical at these points. The 
last equation gives n when I is known.] 

(40) Prove that the Complementary Function of 

becomes negligible when t increases sufficiently, while that of 

d z y d 2 y _ 

oscillates with indefinitely increasing amplitude. 

[An equation of this type holds approximately for the angular 
velocity of the governor of a steam turbine. The first equation corre- 
sponds to a stable motion of revolution, the second to unstable motion 
or " hdyting." See the Appendix to Perry's Steam Engine. ] 

(41) fSgove that the general solution of the simultaneous equations : 

m d £=Ve-He d J, 

dt 2 dt 

m dt*~ Me dt' 
where m, V, H, and e are constants, is 

x = A + B cos {at - a), 

V 

y = ■= t + C + B sin ( cot - a), 

He 

where w = — and A, B, C, a are arbitrary constants. 

Given that — = -^ = x = y=0 when t = 0, show that these reduce to 

x=— -(1 -COS tot), 
ton. 

V . 

y= —jj (cot - sin tot), the equations of a cycloid. 



48 DIFFERENTIAL EQUATIONS 

[These equations give the path of a corpuscle of mass m and charge 
e repelled from a negatively-charged sheet of zinc illuminated with 
ultra-violet light, under a magnetic field H parallel to the surface. V is 
the electric intensity due to the charged surface. By finding ex- 
perimentally the greatest value of x, Sir J. J. Thomson determined 

IV m 

— =, from which the important ratio — is calculated when V and H are 

<*>H e 

known. See Phil. Mag. Vol. 48, p. 547, 1899.] 
(42) Given the simultaneous equations, 

- (1*1 1 ..dH* I x „ 



. (1*1 ' ^(Z 2 /, I 2 



where L v L 2 , M, c v c 2 , E and p are constants, prove that I x is of the 

form a x cos pt + A x cos (mt -a) + B 1 cos (nt - /3), 

and 1 2 of the form 

a 2 cos pt + A 2 cos (mt -a) + B 2 cos (nt - /3), 

E 

where a i = TFa(l -P 2c 2 L 2)> 

EM 3 

<X 2 — , p C]C 2 , 

k denoting the expression 

(L X L 2 - M 2 ) c x c 2 ^ - (L lCl + L 2 c 2 )p 2 + 1 ; 

m and n are certain definite constants ; A 1} B v a and /3 are arbitrary 
constants ; and A 2 is expressible in terms of A x and B 2 in terms 
of A 2 . 

Prove further that m and n are real if L v L 2 , M, c v and c 2 are real 
and positive. 

[These equations give the primary and secondary currents I x and 
7 2 in a transformer when the circuits contain condensers of capacities 
Cj and c 2 . L x and L 2 are the coefficients of self-induction and M that 
of mutual induction. The resistances (which are usually very small) 
have been neglected. E sin pt is the impressed E.M.F. of the primary.] 



CHAPTER IV 
SIMPLE PARTIAL DIFFERENTIAL EQUATIONS 

41. In this chapter we shall consider some of the ways in which 
partial differential equations arise, the construction of simple par- 
ticular solutions, and the formation of more complex solutions from 
infinite series of the particular solutions. We shall also explain the 
application of Fourier's Series, by which we can make these complex 
solutions satisfy given conditions. 

The equations considered include those that occur in problems 
on the conduction of heat, the vibrations of strings, electrostatics 
and gravitation, telephones, electro-magnetic waves, and the 
diffusion of solvents. 

The methods of this chapter are chiefly due to Euler, D'Alembert, 
and Lagrange.* 

42. Elimination of arbitrary functions. In Chapter I. we showed 
how to form ordinary differential equations by the elimination of 
arbitrary constants. Partial differential equations can often be 
formed by the elimination of arbitrary functions. 

Ex. (i). Eliminate the arbitrary functions /and F from 

y=f(x-«l) + F(x + at) (!) 

We get ^ =f\x - at) + F'(x + at) • 

and < ^=f"(x-al) + F"(x + at) (2) 

Similarly -~ — - af'(x - at) + aF'(x + at) 

and yi = a 2 f"{x-at)+a 2 F"(x + at) (3) 

* Joseph Louis Lagrange of Turin (1730-1813), the greatest mathematician of 
the eighteenth century, contributed largely to every branch of Mathematics. He 
created the Calculus of Variations and much of" the subject of Partial Differential 
Equations, and he greatly developed Theoretical Mechanics and Inlinitesimal 
Calculus. 

p.d.b. 49 u 



C*o 



50 DIFFERENTIAL EQUATIONS 

From (2) and (3), g = I g : (4) 

a partial differential equation of the second order.* 
Ex. (ii). Eliminate the arbitrary function / from 

■-/©■ 

dz dz n 
so sc^+w^ =0. 

dx * By 

Examples for solution. 

/ Eliminate the arbitrary functions from the following equations : 
y (1) z=f(x + ay). /(2) z=f(x + iy) + F(x-iy),wheTei*=-l. 

v/ (3) z =f (x cos a + y sin a - at) + F(x cos a + y sin a + at). 
/(4) z=f(x 2 -y 2 ). v (5) z = e ax + b vf(ax-by). 

- (6) -*/©• 

43. Elimination of arbitrary constants. We have seen in 
Chapter I. how to eliminate arbitrary constants by ordinary 
differential equations. This can also be effected by partials. 

Ex. (i). Eliminate A and p from z = Ae pt sin px. 

d 2 z 

— = - p 2 Ae pt sin px, 

d 2 z 
2 = p 2 Ae pt sin px ; 




d! 



dx 2+ dt 2 



Eliminate a, b, and c from 

z = a(x + y) + b(x-y) +abt + c. 

dz 
We get =-=a + b, 

dz 7 
— =a-b, 
ay 

dz , 

=-=00. 

dt 

* Tin's equation holds for the transverse vibrations of a stretched string. 
The most general solution of it is equation (1), which represents two waves 
travelling with spied a, one to the right and the other to the left. 



PARTIAL DIFFERENTIAL EQUATIONS 51 

But (a + o) 2 -(a-&) 2 = 4a&. 

(!)'-(!)'-£• 

Examples for solution. 

Eliminate the arbitrary constants from the following equations : 
<i(l) z= Ae-P 2t cos px. */(2) z = Ae-? 1 cos qx sin ry, where p 2 = q 2 + r l . 
y/{Z) z = ax + (\-a)y + b. (4) z = ax + by + a 2 + b 2 . 

'(5) 2 = (x-a) 2 + (t/-6) 2 . (6) az + b = a 2 x + y. 

44. Special difficulties of partial differential equations. As we have 
already stated in Chapter I., every ordinary differential equation 
of the n th order may be regarded as derived from a solution con- 
taining n arbitrary constants* It might be supposed that every 
partial differential equation of the n th order was similarly derivable 
from a solution containing n arbitrary functions. However, this is 
not true. In general it is impossible to express the eliminant of 
n arbitrary functions as a partial differential equation of order n. 
An equation of a higher order is required, and the result is not 
unique."!* 

In this chapter we shall content ourselves with finding particular 
solutions. By means of these we can solve such problems as most 
commonly arise from physical considerations. J We may console 
ourselves for our inability to find the most general solutions by the 
reflection that in those cases when they have been found it is often 
extremely difficult to apply them to any particular problem. § 

♦It will be shown later (Chap. VI.) that in certain exceptional cases an 
ordinary differential equation admits of Singular Solutions in addition to the 
solution with arbitrary constants. These Singular Solutions are not derivable 
from the ordinary solution by giving the constants particular values, but are of 
quite a different form. 

tSee Edwards' Differential Calculus, Arts. 512 and 513, or Williamson's 
Differential Calculus, Art. 317. 

X The physicist will take it as obvious that every such problem has a solution, 
and moreover that this solution is unique. From the point of view of pure 
mathematics, it is a matter of great difficulty to prove the first of these facts : 
this proof has only been given quite recently by the aid of the Theory of Integral 
Equations (see Heywood and Frechet's L' Equation de Fredholm et ses application.? 
a la Physique Math6matique). The second fact is easily proved by the aid of 
Green's Theorem (see Carslaw's Fourier's Series and Integrals, p. 206). 

§For example, Whittaker has proved that the most general solution of 
Laplace's equation twit t&v 7VV 

3a; 2 dy 2 3z 2 ~ 



V= I f(x cos t + y sin t + iz, t) dt, 



but if we wish to find a solution satisfying certain given conditions on a given 
surface, we generally use a solution in the form of an infinite series. 






52 DIFFERENTIAL EQUATIONS 

45. Simple particular solutions. 

d 2 z 1 dz 
Ex. (i). Consider the equation ,- 2 = - 2 r (which gives the con- 
duction" of heat in one dimension). This equation is linear. Now, in 
the treatment of ordinary linear equations we found exponentials very 
useful. This suggests z = e" u:+nt as a trial solution. Substituting in 
the differential equation, we get 

a 2 * 

which is true if n = m 2 a 2 . 

ThUs e ""'+" l2a2< is a solution. 
Changing the sign of m, e' mx+nfiaH is also a solution. 

Ex. (ii). Find a solution of the same equation that vanishes when 
t= +oo . 

In the previous solutions t occurs in e w2a2t . This increases with t, 
since m 2 a 2 is positive if m and a are real. To make it decrease, put 
m = ip, so that m 2 a 2 = - p 2 a 2 . 

This gives e ipx ~ p2aH as a solution. 

Similarly e - ^ - ^" 2 ' is a solution. 

Hence, as the differential equation is linear, e~ p2a2t (Ae ipx + Be~ ipx ) is 
a solution, which we replace, as usual, by 

e~ p ' kl \E cos px + F sin px). 

d 2 z d 2 z 
Ex. (iii). Find a solution of ^-^ + ^~^=^ which shall vanish when 

y = + oo , and also when x = 0. ^ 

Putting z = e mx+ny , we get (m 2 + n 2 )e mx+n v = 0, so m 2 + n 2 =0. 

The condition when y = + oo demands that n should be real and 
negative, say n= -p. 

Then m = ± ip. 

Hence e~v y {Ae ipx + Be~ ipx ) is a solution, 

i.e. e~ vi '(E cos px.-\- F sm px) is a solution. 

But z = if x = 0, so E = 0. 

The solution required is therefore Fe~ P!/ sin px. 

Examples for solution. 

O^U u it 

(1) ~-| = =-£-, given that y = when x = + oo and also when t = + oo . 

d 2 z 1 9 2 z 

(2) — - =-x =-j, given that z is never infinite (for any real values of 
v ox 2 a 2 oy 2 

x or y), and that 2=0 when x = or y = 0. 

(3) \-a „-=0, given that z is never infinite, and that =- =0 when 

K ' dx dy s ox 

■x=*y=>(X 



PARTIAL DIFFERENTIAL EQUATIONS 53 

d 2 V d 2 V d 2 V 

(4) 5 , 2 +2 _ 2 + "^T =: ^' gi yen th at V=0 when z=+oo, when 

y = - oo , and also when 2 = 0. 

(5) p-g = - - , given that V is never infinite, and that V = C and 

dV dV dV . , 

x— •»-=-= -5-= when x = y = z=0. 

ox ay oz 

d 2 V d 2 V dV 

(6) -2^" + "2~i = ar> given that V =0 when £ = + oo , when x=0 or 

I, and when ?/ = or I. 

46. More complicated initial and boundary conditions.* In Ex. (iii) 
of Art. 45, we found Fe~ py sin fx as a solution of 

dx 2+ dy 2 ~ ' 
satisfying the conditions that z=0 if y= + oo or if £=0. 

Suppose that we impose two extra conditions,"} - say 2=0 if x = l 
and 2 = Ix - x 2 if «/ =0 for all values of x between and I. 
The first condition gives sin pi =0, 

i.e. pi = nw, where n is any integer. 
For simplicity we will at first take I = w, giving p = n, any integer. 
The second condition gives F sin px = irx- x 2 for all values of x 
between and ir. This is impossible. 

However, instead of the solution consisting of a single term, we 
may take 

F x e~ y sin x + F 2 e~ 2y sin 2x + F 3 e- 3y sin 3x + . . . , 

since the equation is linear (if this is not clear, cf. Chap. TIL Art. 25), 
giving p the values 1, 2, 3. ... and adding the results. 

By putting y = and equating to irx - x 2 we get 
F A sin x + F 2 sin 2x + F 3 sin 3x + . . . 
= ttx-x 2 for all values of & between and tr. 

The student will possibly think this equation as impossible to 
satisfy as the other, but it is a remarkable fact that we can choose 
values of the F's that make this true. 

This is a particular case of a more general theorem, which we 
now enunciate. 

* As t usually denotes time and x and y rectangular coordinates, a condition 
such as z = when t = is called an initial condition, while one such as z = if 
a; = 0, or if x = l, or if y = x, is called a boundary condition. 

fThis is the problem of finding the steady distribution of temperature in a 
semi-infinite rectangular strip of metal of breadth I, when the infinite sides are 
kept at 0° and the base at (Ix - x 2 )°. 



54 DIFFERENTIAL EQUATIONS 

47. Fourier's Half-Range Series. Every function of x which 
satisfies certain conditions can be expanded in a convergent series 
of the form 

/ (x) = a 1 sin x + a 2 sin 2x + a 3 sin 3x + . . . to inf. 

for all values of x between and -k (but not necessarily for the 
extreme values x=0 and x = ir). 

This is called Fourier's * half-range sine series. 

The conditions alluded to are satisfied in practically every 
physical problem."]" 

Similarly, under the same conditions f(x) may be expanded in 
a half-range cosine series 

l + Z x cos x + 1 2 cos 2x + 1 3 cos 3x + . . . to inf. 

These are called half-range series as against the series valid 
between and 2ir, which contains both sine and cosine terms. 

The proofs of these theorems are very long and difficult. J How- 
ever, if it be assumed that these expansions are possible, it is easy to 
find the values of the coefficients. 

Multiply the sine series by sin nx, and integrate term by term, § 
giving 

pir PIT rn 

I f(x)smnxdx = a 1 \ sin x sin nx dx + a 2 1 sin 2x sin nxdx + ... . 

Jo Jo Jo 

The term with a n as a factor is 
sin 2 nx dx 



a n I si] 
Jo 



a r 1 • ~\ n 

(1 -cos 2nx)dx = ~ x - „ sin 2 nx\ 
2 L 2/i Jo 



2 Jo 

= ^a n 7r. 

* Jean Baptiste Joseph Fourier of Auxerre (1768-1830) is best known as the 
author of La Th4orie analytique de la chaleur. His series arose in the solution of 
problems on the conduction of heat. 

f It is sufficient forf(x) to be single-valued, finite, and continuous, and have 
only a limited number of maxima and minima between sc = and x = w. However, 
these conditions are not necessary. The necessary and sufficient set of conditions 
has not yet been discovered. 

X For a full discussion of Fourier's Series, see Carslaw's Fourier's Series and 
Integrals and Hobson's Theory of Functions. 

§ The assumption that this is legitimate is another point that requires 
justification. 



PARTIAL DIFFERENTIAL EQUATIONS 65 

The term involving any other coefficient, say a r , is 
a r sin rx sin nx dx 

= -~\ {cos (n - r)x- cos (n+r)x}dx 

^a r r sia(n -r)x sin (n+r)x~y 
2l n-r n + r J = ' 

So all the terms on the right vanish except one. 

Thus I f(x) sin nx dx = \a n ir, 

2 f /., 
01 a n = - \ f(%) sin nx dx. 

Similarly, it is easy to prove that if 

f(x) =b + b 1 cos x + 6 2 cos 2o3 + . . . 
for values of a; between and tt, then 

1 r 

ttJo 
and b n = /(«) cos wee (Zx 

for values of n other than 0. 

48. Examples of Fourier's Series. 

(i) Expand ttx-x 2 in a half-range sine series, valid between x = 
and x = 7r. 

It is better not to quote the formula established in the last article. 
Let 7rx - x 2 = a 1 sin x + a 2 sin 2x + a 3 sin 3x + . . . . 

Multiply by sin nx and integrate from to tt, giving 

I (7rx - x 2 ) sin nx dx = a n \ sin 2 nxdx = ~ a n , as before. 

Now, integrating by parts, 

I (irx-x 2 ) sin nxdx = \ — {ttx - x 2 ) cos nx\ +- (tt - 2x) cos nx dx 
Jo L n J wj v 

= + — = (x - 2x) sin wz + — ~\ sin wa; tZir 
U 2 Jo ™ 2 Jo 

2 r t 4 . 

= — 5 cos nx = — if n is odd or if « is even. 

w L Jo n 

o 

Thus a n = — g if w is odd or if w is even, giving finally 



ttx - x 2 = - (sin cc + -gV sin 3x + T I T sin 5x + . . . ). 

7T 



56 DIFFERENTIAL EQUATIONS 

(ii) Expand /(as) in a half-range series valid from x =0 to x = ir, where 
f{x) — mx between x — and x = — 

Z 

and f(x) = m{7r-x) between x = — and sc = 7r. 

In this case f(x) is given by different analytical expressions in 
different parts of the range.* The only novelty lies in the evaluation 
of the integrals. 

In this case 

1 f(x) sin nx dx= I f(x) sin nx dx+\ f(x) sin wx cZx 
Jo Jo J| 

7T 

= I mx sin nx dx + I m(7r - x) sin wx dx. 
Jo J| 

We leave the rest of the work to the student. The result is 

d-fyy 

— (sin z-^sin Sx + ^ r sin So;-^ 8 * 11 7as + ...). 

The student should draw the graph of the given function, and 
compare it with the graph of the first term and of the sum of the first 
two terms of this expansion. f 

Examples for solution. 

Expand the following functions in half -range sine series, valid 
between x = and x = tt : 

(1) 1. (2) x. (3) X s . (4) cos a. (5) e x . 

(6) f(x)=0 from x=0 to x =—, and from x=— to tt, 

f(x) = (4:X-7r)(37r-4:x) from x = — to x=-r-. 

(7) Which of these expansions hold good (a) for x — ? 

(6) for x — 7t ? 

49. Application of Fourier's series to satisfy boundary conditions. 

We can now complete the solution of the problem of Art. 46. 
We found in Art. 46 that 

F x e-v sin x+F 2 e~ 2 " sin 2x +_F 3 e~ 3 " sin 3a; + ... 
satisfied all the conditions, if 

F x sin x + F 2 sin 2x + F 3 sin Sx + . . . = irx - x 2 
for all values of x between and 7r. 

* Fourier's theorem applies even if f(x) is given by a graph with no analytical 
expression at all, if the conditions given in the footnote to Art. 47 are satisfied. 

For a function given graphically, these integrals are determined by arith- 
metical approximation or by an instrument known as a Harmonic Analyser. 

t Several of the graphs will be found in Carslaw's Fonrur's Series mid Inter/rah, 
Art. 59. More elaborate ones are given in the Phil. May., Vol. 45 (1898). 



PARTIAL DIFFERENTIAL EQUATIONS 57 

In Ex. (i) of Art. 48 we found that, between and -k, 

o 

r- (sin x + -jV sm 3a; + y^ T sin 5a; + ...) = ttx - a; 2 . 

7T 

Thus the solution required is 

- {e~' J sin a; + ^ T e~ Z!l sin 3a; + 1-J--5 e~ 5j/ sin 5a; + . . .). 

7T 

50. In the case when the boundary condition involved I instead 
of ir, we found Fe~ vy sin jpx as a solution of the differential equation, 
and the conditions showed that p, instead of being a positive integer 
n, must be of the form mr/l. 

Thus F^e-^' sin ttx/1 + F 2 e~ in ^ 1 sin 2 irx/l + . . . 

satisfies all the conditions if 

F t sin 7rx{l + F 2 sin Sirx/l + ...=lx-x 2 
for all values of x between and I. 

I 2 I 2 

Put ttx/1 = z. Then Ix -x 2 = — -A-kz-z 2 ). The F's are thus -= 

7T 7T 

times as much as before. The solution is therefore 

8l 2 

" -3 ( e_Tr?// ' sin 7ra; /^ + irr e ~' inv " sm STrai/Z + xi^ 6 " 5 ^' sin 57r:r /^ + •••)• 

MISCELLANEOUS EXAMPLES ON CHAPTER IV. 

1 -Jt 

(1) Verify that V = — re 4^< is a solution of 

3 2 V_1 dV 
Bx 2 ~ K dt ' 

(2) Eliminate A and p from V =Ae~ px sin (2p 2 Kt-px). 

dV d 2 V 

(3) Transform -^- = K ^ - hV 

3W „d 2 W 
t0 . -U= K ~W 

by putting V = e- ht W. 

[The first equation gives the temperature of a conducting rod whose 
surface is allowed to radiate heat into air at temperature zero. The 
given transformation reduces the problem to one without radiation.] 

(4) Transform 

dV_Klf 2 dV\ dW d 2 W 

dt~r 2 dr\ dr) dt dr 2 

by putting W = rV. 

[The first equation gives the temperature of a sphere, when heat 
flows radially.] 



58 DIFFERENTIAL EQUATIONS 

(5) Eliminate the arbitrary functions from 

vJ-[f(r-at) + F(r + at)]. 

(6) (i) Show that if e mx+int is a solution of 

dV d 2 V 

where n and h are real, then m must be complex. 

(ii) Hence, putting m=-g-if, show that V e- ffx sin (nt -fx) is a 
solution that reduces to F sin nt for x = 0, provided K(g 2 -f 2 )=h and 
n = 2Kfg. 

(iii) If V=0 when x= +oo , show that if K and n are positive so 
are g and /. 

[In Angstrom's method of measuring K (the " diffusivity "), one 
end of a very long bar is subjected to a periodic change of temperature 
V sin nt. This causes heat waves to travel along the bar. By measur- 
ing their velocity and rate of decay n/f and g are found. K is then 
calculated from K = n/2fg.] 

dV d 2 V 

(7) Find a solution of -^- = K^~y reducing to V sin nt for x=0 

and to zero for x = + oo . dt dx 

[This is the problem of the last question when no radiation takes 
place. The bar may be replaced by a semi-infinite solid bounded by 
a plane face, if the flow is always perpendicular to that face. Kelvin 
found K for the earth by this method.] 

(8) Prove that the simultaneous equations 

are satisfied by V = V e-^ +i J^ +int , 

if g*-p = RK-n*LC, 

2fg = n(RC + LK), 

and V( R + iLn) = V 2 (K + iCn). 

[These are Heaviside's equations for a telephone cable with resist- 
ance R, capacity C, inductance L, and leakance K, all measured per 
unit length. / is the current and V the electromotive force.] 

(9) Show that in the last question g is independent of n if RC = KL. 
[The attenuation of the wave depends upon g, which in general 

depends upon n. Thus, if a sound is composed of harmonic waves of 
different frequencies, these waves are transmitted with different degrees 
of attenuation. The sound received at the other end is therefore 



MISCELLANEOUS EXAMPLES 59 

distorted. Heaviside's device of increasing L and K to make RC = KL 
prevents this distortion.] 

(10) In question (8), if L = K=0, show that both V and / are 
propagated with velocity <\/(2n/RC). 

[The velocity is given by n/f.] 

(11) Show that the simultaneous equations 





kdPdy 5/3. 


fidadR dQ. 




c dt dy dz ' 


c dt dy dz ' 




k dQ da dy 


lxd($_dP dR, 


* 


c dt dz dx ' 


c dt dz dx 




kdR d(3 da 
c dt dx dy ' 


lxdy_dQ dP, 

c dt dx dy ' 


are satisfied by 


P=0; 


a=0; 




£=0; 


/3 = /3 sin y(x-vt) ; 




R=R sin p (x - 


vt); y = 0; 


provided that v 


= c/Vk/uL and /5 = 


= -VWfx)K 



[These are Maxwell's electromagnetic equations for a dielectric of 
specific inductive capacity k and permeability /x. P, Q, R are the 
components of the electric intensity and a, /3, y those of the magnetic 
intensity, c is the ratio of the electromagnetic to the electrostatic 
units (which is equal to the velocity of light in free ether). The solution 
shows that plane electromagnetic waves travel with the velocity c/^/k/u, 
and that the electric and magnetic intensities are perpendicular to the 
direction of propagation and to each other.] 

dV d 2 V 

(12) Find a solution of -^- — K ^-j su ch that 

F=/=oo if t= +oo ; 

F=0 if x=0 or ir, for all values of t ; 

V = ttx-x 2 if t = 0, for values of x between and ir. 
[N.B. Before attempting this question read again Arts. 46 and 49. 
V is the temperature of a non-radiating rod of length ir whose ends are 
kept at 0°, the temperature of the rod being initially (ttx-x 2 )° at a 
distance x from an end.] 

(13) What does the solution of the last question become if the 
length of the rod is I instead of 7r ? 

[N.B. Proceed as in Art. 50.] 

(14) Solve question (12) if the condition 7 = for x = or ir is 

dV 
replaced by •=— = for x = or ir. 

[Instead of the ends being at a constant temperature, they are here 
treated so that no heat can pass through them.] 

(15) Solve question (12) if the expression ttx-x 2 is replaced by 100. 



60 DIFFERENTIAL EQUATIONS 

dV d 2 V 

(16) Find a solution of -=-=K =-— such that 

at ox* 

T^=oo if t= +00 ; 

F = 100 if x=0 or ir for all values of t ; 
F = if /=0 for all values of x between and x. 
[Here the initially ice-cold rod has its ends in boiling water.] 

(17) Solve question (15) if the length is I instead of ir. If I increases 
indefinitely, show that the infinite series becomes the integral 



200 r 1 

7r J a 



e KaH sin ax da. 



[N.B. This is called a Fourier's Integral. To obtain this residt 
put (2r + l)Tr/l = a and 2x// = ^a. 

Kelvin used an integral in his celebrated estimate of the age of the 
earth from the observed rate of increase of temperature underground. 
(See example (107) of the miscellaneous set at the end of the book.) 
Strutt's recent discovery that heat is continually generated within the 
earth by radio-active processes shows that Kelvin's estimate was too 
small.] 

dV d 2 V 

(18) Find a solution of -=-==K^~y such that 

V is finite when t = + co ; 

dV ■] 

-=-=0 when x = 0, 1 
ox Y for all values of t ; 

F=0 when x = lj 

V=V when t = 0, for all values of x between and I. 

[If a small test-tube containing a solution of salt is completely 

submerged in a very large vessel full of water, the salt diffuses up out 

of the test-tube into the water of the large vessel. If V Q is the initial 

concentration of the salt and I the length of test-tube it fills, V gives 

the concentration at any time at a height x above the bottom of the 

dV 
test-tube. The condition ^— = when x = means that no diffusion 

ox 

takes place at the closed end. V = when x = l means that at the top 

of the test-tube we have nearly pure water.] 

(19) Find a solution of ^y = v 2 ^~ such that 

y involves x trigonomctrically ; 

?/=0 when x — or it, for all values of t ; 

dy 

~=0 when 1 = 0, for all values of x ; 

ot 



y = mx between # = and — , 

y = ?n{TT-x) between £=o and 7r, 



for nil values of I. 



MISCELLANEOUS EXAMPLES 61 

[N.B. See the second worked example of Art. 48. 

y is the transverse displacement of a string stretched between two 
points a distance ir apart. The string is plucked aside a distance 
nnr/2 at its middle point and then released.] 

UtU 

* (20) Writing the solution of j~ = D 2 y, where D is a constant, in 

the form 



d 2 y d 2 y 
deduce the solution of ^ri=-^4 in the form 
ox 2 ol 2 



y = e xD A+e- xD B, 

=tHt in the form 
at 2 

y=f(t + x) + F(t-x) 



by substituting =- for D, f(t) and F(t) for A and B respectively, and 

using Taylor's theorem in its symbolical form 

f(t + x) = e* D f{t). 

[The results obtained by these symbolical methods should be 
regarded merely as probably correct. Unless they can be verified by 
other means, a very careful examination of the argument is necessary 
to see if it can be taken backwards from the result to the differential 
equation. 

Heaviside has used symbolical methods to solve some otherwise 
insoluble problems. See his Electromagnetic Theory. J 

Q/1J 

* (21) From the solution of -— = Dhi, where D is a constant, deduce 

. . ay o £ y . . . 
that of jr- — %fir in the form 

d 2 f x 2 d i f 

[This is not a solution unless the series is convergent.] 
Use this form to obtain a solution which is rational, integral, and 
algebraic of the second degree in t. 

d 2 y d 2 y 
*(22) Transform the equation ^j — ^^i by changing the inde- 
pendent variables x and i to Z and T, where 
X = x-at; T = x + at. 
Hence solve the original equation. 

*To be omitted on a first reading. 



CHAPTER V 

EQUATIONS OF THE FIEST ORDER BUT NOT OF THE 
FIRST DEGREE 

51. In this chapter we shall deal with some special typee of 
equations of the first order and of degree higher than the first for 
which the solution can sometimes be obtained without the use of 
infinite series. 

These special types are : 

(a) Those solvable for p. 

(b) Those solvable for y. 

(c) Those solvable for x. 

52. Equations solvable for p. If we can solve for p, the equation 
of the n th degree is reduced to n equations of the first degree, to 
which we apply the methods of Chap. II. 

Ex. (i). The equation p 2 + px+py + xy = gives 
p= -x or p= -y ; 
from which 2y = - x 2 + c 1 or x = - log y + c 2 ; 

or, expressed as one equation, 

(2y + x 2 - Cl )(x + \ogy-c 2 )=0 (1) 

At this point we meet with a difficulty ; the complete primitive 
apparently contains two arbitrary constants, whereas we expect only 
one, as the equation is of the first order. 

But consider the solution 

(2y + x 2 - c)(x + log y - c) = (2) 

If we are considering only one value of each of the constants c, c lt 
and c 2 , these equations each represent a pair of curves, and of course 
not the same pair (unless c = c 1 =c 2 ). But if we consider the infinite 
set of pairs of curves obtained by giving the constants all possible 
values from - oo to + go , we shall get the same infinite set when taken 
altogether, though possibly in a different order. Thus (2) can be taken 
as the complete primitive. 

62 



EQUATIONS OF THE FIRST ORDER 63 

Ex. (ii). p 2 +p -2=0. 

Here p = 1 or p = - 2, 

giving y=x + c 1 or y=-2x + c 2 . 

As before, we take the complete primitive as 
(y-x-c)(y + 2x-c)=0, 
not (y-x-c 1 )(y + 2x-c 2 )=0. 

Each of these equations represents all lines parallel either to 
y=x or to y = -2x. 

Examples for solution. 
♦/(I) p 2 + p-6=0. ^(2) p 2 + 2xp = 3x 2 . v/(3) p 2 = a .B 

(4) x + yp 2 =p(l+xy). (5) p 3 -p(x 2 + xy + y 2 )+xy(x + y)=0. 

v-<6) y 2 - 2p cosh x + 1=0. 
53. Equations solvable for y. If the equation is solvable for y, 
we differentiate the solved form with respect to x. 
Ex. (i). p 2 -py + x=>0. 

Solving for y, V=P + ~- 

„.„ . . dp 1 x dp 

Differentiating, P^ir ^ 5 j~ » 

° r dx p p l ax 

l\dx x 

p) dp p 2 

This is a linear equation of the first order, considering p as the 

independent variable. Proceeding as in Art. 19, the student will obtain 

x=p(c+cosh~ 1 p)(p 2 -l) . 

00 —It 

Hence, as y=p + -, y =* p + (c + cosh^p) (p 2 - 1) . 

These two equations for x and y in terms of p give the parametric 
equations of the solution of the differential equation. For any given 
value of c, to each value of p correspond one definite value of x and 
one of y, defining a point. As p varies, the point moves, tracing out 
a curve. In this example we can eliminate p and get the equation con- 
necting x and y, but for tracing the curve the parametric forms are as 
good, if not better. 

Ex. (ii). 3p 5 -py + l=0. 

Solving for y, y = 3p* + p~ l . 

Differentiating, p = I2p s -j- - p~ 2 .~, 

i.e. dx = (12p 2 -p' 3 )dp. 

Integrating, x = 4^ 3 + \p~ 2 + c, ~\ 

and from above, y = 3p i + p~ 1 . J 

The student should trace the graph of this for some particular value 
of c, say c = 0. 



64 DIFFERENTIAL EQUATIONS 

54. Equations solvable for x. If the equation is solvable for x, 

we differentiate the solved form with respect to y, and rewrite -y- 

1 
in the form - . 
V 

Ex. p 2 -py + x=0. This was solved in the last article by solving 
for y. 

Solving for x, x=py- p % . 

Differentiating with respect to y, 

1 dp dp 

which is a linear equation of the first order, considering p as the inde- 
pendent and y as the dependent variable. This may be solved as in 
Art. 19. The student will obtain the result found in the last article. 

Examples for solution. 

(1) x = 4p + 4:p 3 . (2) p 2 -2xp + l=0. 

\ (3) y=p 2 x + p. (4) y=x+p z . 

(5) p 3 + p = ev. (6) 2y+p 2 + 2p = 2x(p + l). 

(7) p 3 -p (y + 3)+x = 0. (8) y = pBm p + cosp. - 

(9) y=p tan 2> + log cos p. (10) e p ~ y =p 2 -l. 

(12) Prove that all curves of the family given by the solution of 
Ex. 1 cut the axis of y at right angles. Find the value of c for that 
curve of the family that goes through the point (0, 1). 

Trace this curve on squared paper. 

(13) Trace the curve given by the solution of Ex. 9 with c=0. 
Draw the tangents at the points given by p = 0, p=l, p = 2 and p = S, 
and verify, by measurement, that the gradients of these tangents are 
respectively 0, 1, 2 and 3. 



CHAPTER VI 
SINGULAR SOLUTIONS* 

55. We know from coordinate geometry that the straight line 

y = mx + — touches the parabola y 2 = iax, whatever the value of m. 

Consider the point of contact P of any particular tangent. At 
P the tangent and parabola have the same direction, so they have 

a common value of -,-, as well as of x and y. 




Fig. 7. 
But for the tangent m=^-=jp say, so the tangent satisfies the 

differential equation y=px+~. 

Hence the equation holds also for the parabola at P, where x, 
y, and p are the same as for the tangent. As P may be any point 
on the parabola, the equation of the parabola y 2 = iax must be a 
solution of the differential equation, as the student will easily verify. 

* The arguments of this chapter will be based upon geometrical intuition. The 
results therefore cannot be considered to be proved, but merely suggested as 
probably true in certain cases. The analytical theory presents grave difficulties 
(see M. J. M. Hill, Proc. Loud. Math. Soc, 1918). 

P.U.E. t)5 B 



66 DIFFERENTIAL EQUATIONS 

In general, if we have any singly infinite system of curves which 
all touch a fixed curve, which we will call their envelope* and if this 
family represents the complete primitive of a certain differential 
equation of the first order, then the envelope represents a solution 
of the differential equation. For at every point of the envelope 
x, y, and p have the same value for the envelope and the curve of 
the family that touches it there. 

Such a solution is called a Singular Solution. It does not 
contain any arbitrary constant, and is not deducible from the 
Complete Primitive by giving a particular value to the arbitrary 
constant in it. 

Example for solution. 

Prove that the straight line y = x is the envelope of the family of 
parabolas y = x + \{x-c) 2 . Prove that the point of contact is (c, c), 
and that p — \ for the parabola and envelope at this point. Obtain 
the differential equation of the family of parabolas in the form 
y — x + (p- 1) 2 , and verify that the equation of the envelope satisfies this. 

Trace the envelope and a few parabolas of the family, taking c as 
0, 1, 2, etc. 

56. We shall now consider how to obtain singular solutions. It 
has been shown that the envelope of the curves represented by the 
complete primitive gives a singular solution, so we shall commence 
by examining the method of finding envelopes. 

The general method t is to eliminate the parameter c between 
f(x, y, c) =0, the equation of the family of curves, and 

I- 

E.g. if f(x,y,c) = is y-cx--=0, (1) 

|=0 is - * + l=0 (2) 

giving c = ± l/^x. 

*In Lamb's Infinitesimal Calculus, 2nd ed., Art. 155, the envelope of a 
family is defined as the locus of ultimate intersection of consecutive curves of 
the family. As thus defined it may include node- or cusp loci in addition to or 
instead of what wo have called envelopes. (We shall give a geometrical reason for 
this in Art. 56 ; see Lamb for an analytical proof.) 

t See Lamb's Infinitesimal Calculus, 2nd ed., Art. 155. If f(x, y, c) is of 
the form Lc i + Mc + N, the result comes to J\1 2 = 4LN. Tims, for 

1 o 

y - ex - - = 0, 

9 c ' 

i. e. c 2 x - cy + 1 = 0, 
the result is y 2 = 4x. 



SINGULAR SOLUTIONS 



67 



Substituting in (1), 



y=±2^/x, 
or y 2 = 4:X. 

This method is equivalent to finding the locus of intersection of 
f(x, y, c)=0, 
and f(x,y,c + h)=0, 

two curves of the family with parameters that differ by a small 
quantity h, and proceeding to the limit when h approaches zero. 
The result is called the c-discriminant oif(x, y, c) =0. 

57. Now consider the diagrams 8, 9, 10, 11. 
Fig. 8 shows the case where the curves of the family have 
no special singularity. The locus of the ultimate intersections 




Fig 8. 



PQRSTUV is a curve which has two points in common with each 
of the curves of the family (e.g. Q and R lie on the locus and also 
on the curve marked 2). In the limit the locus PQRSTUV there- 
fore touches each curve of the family, and is what we have defined 
as the envelope. 

In Fig. 9 each curve of the family has a node. Two con- 
secutive curves intersect in three points (e.g. curves 2 and 3 in the 
points P, Q, and R). 

The locus of such points consists of three distinct parts EE', 
AA', and BB' . 

When we proceed to the limit, taking the consecutive curves 
ever closer and closer, A A' and BB' will move up to coincidence 
with the node-locus iVA T ', while EE' will become an envelope. So 



68 



DIFFERENTIAL EQUATIONS 



in this case we expect the c-discriminant to contain the square of 
the equation of the node-locus, as well as the equation of the envelope. 



E!-^£-»vv«i 




Fig. 9. 

As Fig. 10 shows, the direction of the node-locus NN' at any 
point P on it is in general not the same as that of either branch of 
the curve with the node at P. The node-locus has x and y in common 
with the curve at P, but not p, so the node-locus is not a solution of 
the differential equation of the curves of the family. 




Fig. 10. 

If the node shrinks into a cusp, the loci EE' and NN' of Fig. 10 
move up to coincidence, forming the cusp-locus CC of Fig. 11. 
Now NN' was shown to be the coincidence of the two loci AA' and 
BB' of Fig. 9, so CC is really the coincidence of three loci, and 
its equation must be expected to occur cubed in the c-discriminant. 

Fig. 11 shows that the cusp-locus, like the node-locus, is not 
(in general) a solution of the differential equation. 




K 



FIG. 11. 



To sum up, we may expect the c-discriminant to contain 
(i) the envelope, 
(ii) the node-locus squared, 
(iii).« the cusp-locus cubed. 



SINGULAR SOLUTIONS 



69 



The envelope is a singular solution, but the node- and cusp- 
loci are not (in general *) solutions at all. 



58. The following examples will illustrate the preceding results : 

Ex. (i). y=p 2 . 

The complete primitive is easily found to be iy = (x-c) 2 , 
i.e. c 2 -2cx + x 2 -±y = 0. 

As this is a quadratic in c, we can write down the discriminant at 
once as (2z) 2 = 4(a; 2 -4*/), 

i.e. y = 0, representing the envelope of the family of equal parabolas 
given by the complete primitive, and occurring to the first degree only, 
as an envelope should. 

y 




Flu. 12. 



Ex. (ii). 



%y = 2px-2 



V 



Proceeding as in the last chapter, we get 



i.e. px 2 - 2p 2 = (2x* - ipx) 



dp 
dx' 



i.e. a; 2 -2p = or p = 2x 



dp 
dx' 



.(A) 



dx dp 

— =2 > 
x p 

* We say in general, because it is conceivable that in some special example a 
node- or cusp-locus may coincide with an envelope or with a curve of the family. 



70 



DIFFERENTIAL EQUATIONS 



log x = 2 log p - log c, 
cx=p 2 , 

1 8 

whence 3y = 2c%* - 2c, 

i.e. (3?/ + 2c) 2 = 402?, a family of semi-cubical parabolas with their cusps 

on the axis of y. 

The c-discriminant is (3y - x 3 ) 2 = 9y 2 , 
i.e. x 3 (6?/-x 3 )=0. 

The cusp-locus appears cubed, and the other factor represents the 
envelope. 

It is easily verified that 6y = x 3 is a solution of the differential 
equation, while x=0 (giving p = oo) is not. 

If we take the first alternative of the equations (a), 



i.e. x< 



2p=0, 



we get by substitution for p in the differential equation 

3*/ =4*3, 
i.e. the envelope. 

This illustrates another method of finding singular solutions. 




Examples for solution. 

Find the complete primitives and singular solutions (if any) of the 
following differential equations. Trace the graphs for Examples 1-4: 
(1) ±p 2 -9x = 0. (2) ip 2 (x-2) = l. 



(3) xp 2 -2yp + 4x = 0. 
(5) p 2 + 2xp-y = 0. 
(7) ixp 2 + iyp - 1 =0. 



(4) ? ) 2 + ?/ 2 -l=0. 
(6) xp 2 -2yp + l=0. 



SINGULAR SOLUTIONS 71 

59. The p-discriminant. We shall now consider how to obtain 
the singular solutions of a differential equation directly from the 
equation itself, without having to find the complete primitive. 

Consider the equation x 2 p 2 - yp + 1 =0. 

If we give x and y any definite numerical values, we get a quad- 
ratic for p. For example, if 

a = v % y=3, 2^ 2 -3^ + l=0, 
p=\ or 1. 

Thus there are two curves of the family satisfying this equation 
through every point. These two curves will have the same tangent 
at all points where the equation has equal roots in p, i.e. where 
the discriminant y 2 - 4x 2 =0. 

Similar conclusions hold for the quadratic Lp 2 + Mp+N=0 } 
where L, M, N are any functions of x and y. There are two curves 
through every point in the plane, but these curves have the same 
direction at all points on the locus M 2 - 4LN =0. 

More generally, the differential equation 

f(x, y, p) = L p" +L lP "-i +L 2 p- 2 + ... +L n =0, 
where the L's are functions of x and y, gives n values of p for a 
given pair of values of x and y, corresponding to n curves through 
any point. Two of these n curves have the same tangent at all 
points on the locus given by eliminating p from 

J{x,y,p)=o, 

l -»■ 

for this is the condition given in books on theory of equations for 
the existence of a repeated root. 

We are thus led to the ^-discriminant, and we must now in- 
vestigate the properties of the loci represented by it. 

60. The Envelope. The ^-discriminant of the equation 

1 

y=px + ~ 

or p 2 x-py + 1=0 f*- 

is y 2 = ix. 

We have already found that the complete primitive consists of 
the tangents to the parabola, which is the singular solution. Two 
of these tangents pass through every point P in the plane, and 
these tangents coincide for points on the envelope. 



72 



DIFFERENTIAL EQUATIONS 



This is an example of the 59-discriminant representing an envelope. 
Fig. 15 shows a more general case of this. 




FIG. 14. 



Consider the curve SQP as moving up to coincidence with the 
curve PRT, always remaining in contact with the envelope QRU. 
The point P will move up towards R, and the tangents to the two 
curves through P will finally coincide with each other and with the 
tangent at the envelope at R. Thus R is a point for which the p'a 
of the two curves of the system through the point coincide, and 
consequently the ^-discriminant vanishes. 

U 




Fig. 15. 

Thus the jo-discriminant may be an envelope of the curves of 
the system, and if so, as shoAvn in Art. 55, is a singular solution. 

61. The tac-locus. The envelope is thus the locus of points 
where two consecutive curves of the family have the same value 
of p. But it is quite possible for two non-consecutive curves to 
touch. 

Consider a family of circles, all of equal radius, whose centres 
lie on a straight line. 



SINGULAR SOLUTIONS 



73 



Fig. 16 shows that the line of centres is the locus of the point 
of contact of pairs of circles. This is called a,, tac-locus. Fig. 17 




fig. 16. 



shows circles which do not quite touch, but cut in pairs of neigh- 
bouring points, lying on two neighbouring loci AA' , BB' . When, 
we proceed to the limiting case of contact these two loci coincide 
in the tac-locus TT . Thus the ^-discriminant may be expected to 
contain the equation of the tac-locus squared. 




Fig. 17. 

It is obvious that at the point P in Fig. 16 the direction of 
the tac-locus is not the direction of the two circles. Thus the 
relation between x, y, and p satisfied by the circles will not be 
satisfied by the tac-locus, which has the same x and y but a different 
p at P. In general, the tac-locus does not furnish a solution of tlie 
differential equation. 

62. The circles of the last article are represented by 
(x + c) 2 +y 2 =r 2 , 
if the line of centres is Ox. 

This gives x+c = Vr 2 - y 2 , 

or 1= -yp/Vr 2 -y 2 , 



i.e. 



y 2 p 2 + y 2 



r 2 =0. 



The ^-discriminant of this is y 2 (y 2 -r 2 ) = 0. 

The line y=0 (occurring squared, as we expected) is the tac- 
locus, y=±r are the envelopes EE' and FF' of Fig. 16; y = ±r, 
giving p=0, are singular solutions of the differential equation, but 
y = does not satisfy it. 

63. The cusp-locus. The contact that gives rise to the equal 
roots in p may be between two branches of the same curve instead 



74 



DIFFERENTIAL EQUATIONS 



of between two different curves, i.e. the ^-discriminant vanishes at 
a cusp. 

As shown in Fig. 18, the direction of the cusp-locus at any 
point P on it is in general not the same as that of the tangent to 
the cusp, so the cusp-locus is not a solution of the differential equation. 

C' 




Fig. 18. 



It is natural to enquire if the equation of the cusp-locus will 
appear cubed in the p- discriminant, as in the c-discriminant. To 
decide this, consider the locus of points for which the two p's are 
nearly but not quite equal, when the curves have very flat nodes. 
This will be the locus NN' of Fig. 19. In the limit, when the nodes 




Fig. 19. 



contract into cusps, we get the cusp-locus, and as in this case there 
is no question of two or more loci coinciding, we expect the p- 
discriminant to contain the equation of the cusp-locus to the first 
power only. 

64. Summary of results. The ^-discriminant therefore may be 

expected to contain 

(i) the envelope, 

(ii) the tac-locus squared, 

4 (iii) the cusp-locus, 

and the c-discriminant to contain 

(i) the envelope, 

(ii) the node-locus squared, 

(iii) the cusp-locus cubed. 



SINGULAR SOLUTIONS 



75 



Of these only the envelope is a solution of the differential 
equation. 

65. Examples. 

Ex.(i). ? 2(2-3 <V ) 2 = 4(l-t/). 

Writing this in the form 

fa_ 2-3y 

dy- ± 2V(i-y)' 

we easily find the complete primitive in the form 

(a>-c) 8 =y a (l-y). 
The c-discriminant and ^-discriminant are respectively 
y 2 (l-y)=0 and (2-3y) 2 (l -y)=0. 

1 - y=0, which occurs in both to the first degree, gives an envelope ; 
y=0, which occurs squared in the c-discriminant and not at all in 
the p-discriminant, gives a node-locus ; 2 - Sy = 0, which occurs squared 
in the ^-discriminant and not at all in the c-discriminant, gives a 
tac -locus. 

It is easily verified that of these three loci only the equation of the 
envelope satisfies the differential equation. 




no. 20. 

Ex. (ii). Consider the family of circles 

x 2 + y 2 + 2cx + 2c 2 -l=0. 
By eliminating c (by the methods of Chap. I.), we obtain the differ- 
ential equation 

2y 2 p 2 + 2xyp + x 2 + y 2 - 1 =0. 



76 



DIFFERENTIAL EQUATIONS 



The c- and ^-discriminants are respectively 

x 2 -2(x 2 + y 2 -l)=0 and x 2 y 2 -2y 2 {x 2 + y 2 -l)=0, 
i.e. x 2 + 2y 2 -2=0 and y 2 (x 2 + 2y 2 -2)=0. 

x 2 + 2y 2 -2=0 gives an envelope as it occurs to the first degree in 
both discriminants, while y = gives a tac-locus, as it occurs squared 
in the p-discriminant and not at all in the c-discriminant. 




FIG. 2t. 

Examples for solution. 

In the following examples find the complete primitive if the differ- 
ential equation is given or the differential equation if the complete 
primitive is given. Find the singular solutions (if any). Trace the 
graphs. 

(1) ix(x-l){x-2)p 2 -{3x 2 -6x + 2) 2 = 0. (2) 4*;> 2 -(3z-l) 2 =0. 

(3) yp 2 -2xp + y = 0. (4) 3xp 2 -6yp + x + 2y=0. 

(5) p 2 + 2px 3 -ix 2 y = 0. '6) p 3 -±xyp + 8y 2 =0. 

(7) x 2 + ?/ 2 -2c£ + c 2 cos 2 a = 0. (8) c 2 + 2cy -x 2 + \ = 0. 

(9) c 2 + (x + y)c + l-xy = 0. (10) x 2 + y 2 + 2cxy + c 2 - 1 =0. 

66. Clairaut's Form.* We commenced this chapter by con- 
sidering the equation 



y = px + 



V 



♦Alexis Claude Clairaut, (if Paris (1713-176,")), although best known in con- 
nection with differential equations, wrote chiefly on astronomy. 



SINGULAR SOLUTIONS 



77 



This is a particular case of Clairaut's Form 

y=px+f(p) (1) 

To solve, differentiate with respect to y,-*. 



p=p + {x+f'(p)}-£; 



therefore 



dp 



= 0, 



(2) 



dx~ v > P = C 
or 0=x+f'(p) (3) 

Using (1) and (2) we get the complete primitive, the family of 
straight lines, y = cx+f(c) (4) 

If we eliminate p from (1) and (3) we shall simply get the jo-dis- 
criminant. 

To find the c-discriminant we eliminate e from (4) and the result 
of differentiating (4) partially with respect to c, i.e. 

0=x+f(c) (5) 

Equations (4) and (5) differ from (1) and (3) only in having c 
instead of p. The eliminants are therefore the same. Thus both 
disoriminants must represent the envelope. 

Of course it is obvious that a family of straight lines cannot 
have node-, cusp-, or tac-loci. 

Equation (4) gives the important result that the complete primi- 
tive of a differential equation of Clairaut's Form may be written down 
immediately by simply writing c in place of p. 

67. Example. 

Find the curve such that OT varies as tan \fs, where T is the point 
in which the tangent at any point cuts the axis of x, \Js is its inclination 
to this axis, and is the origin. 

y 




O T 

I'm. 22. 



therefore x-- = kp, •< 






78 DIFFERENTIAL EQUATIONS 

From the figure, OT = ON-TN 

= x-y cot \fr 

V 
V 
since tani//-=p; 

y 

V 

i.e. y=px-kp 2 . 
This is of Clairaut's Form, so the complete primitive is 
y = cx- kc 2 , 
and the singular solution is the discriminant of this, 
i.e. x 2 = iky. 

The curve required is the parabola represented by this singular 
solution. The complete primitive represents the family of straight 
lines tangent to this parabola. 

Examples for solution. 

Find the complete primitive and singular solutions of the following 
differential equations. Trace the graphs for Examples (1), (2), (4), (7), 
(8) and (9). 

7 (1) y=px+p 2 . y (2) y=px + p 3 . 

-/ (3) y=px + cosp. (4) y = px + ^{a 2 p 2 + b 2 ). 

^(5) p=log(px-y). (6) sinpxeos y = coapxsiny+p. 

(7) Find the differential equation of the curve such that the tangent 
makes with the co-ordinate axes a triangle of constant area k 2 , and 
hence find the equation of the curve in integral form. 

(8) Find the curve such that the tangent cuts off intercepts from 
the axes whose sum is constant. 

(9) Find the curve such that the part of the tangent intercepted 
between the axes is of constant length. 



MISCELLANEOUS EXAMPLES ON CHAPTER VI. 

Illustrate the solutions by a graph whenever possible. 

(1) Examine for singular solutions p 2 + 2xp = 3x 2 . 

(2) Reduce xyp 2 -(x 2 + y 2 -l)p + xy = 

to Clairaut's form by the substitution X — x 2 ; Y = y 2 . 

Hence show that the equation represents a family of conies touching 
the four sides of a square. 



MISCELLANEOUS EXAMPLES 79 

(3) Show that xyp 2 + (x 2 -y 2 -h 2 )p-xy = 

represents a family of confocal conies, with the foci at (± h, 0), touching 
the four imaginary lines joining the foci to the circular points at infinity. 

(4) Show by geometrical reasoning or otherwise that the sub- 
stitution x = aX+bY, y = a'X+b'Y, 

converts any differential equation of Clairaut's form to another equation 
of Clairaut's form. 

(5) Show that the complete primitive of 8p 3 x = y(12p 2 -9) is 
(x + c) 3 = 3y 2 c, the p-discriminant y 2 (9x 2 -iy 2 )=0, and the c-dis- 
criminant y*(9x 2 - 4ty 2 ) = 0. Interpret these discriminants. 

(6) Reduce the differential equation 

x 2 p 2 + yp(2x + y)+y 2 =0, where p = -r- 

to Clairaut's form by the substitution £=y, rj = xy. 

Hence, or otherwise, solve the equation. 

Prove that y + ix = is a singular solution ; and that y=0 is both 
part of the envelope and part of an ordinary solution. [London.] 

(7) Solve y^iy-Xjj^^ij) » which can be transformed to 
Clairaut's form by suitable substitutions. [London.] 

(8) Integrate the differential equations : 

(i) 3{p + x) 2 = {p-xf. 
(ii) y 2 {\ +ip 2 )- 2pxy -1=0. 
In (ii) find the singular solution and explain the significance of any 
factors that occur. [London.] 

(9) Show that the curves of the family 

y 2 -2cx 2 y + c 2 (x i -x 3 )=0 

all have a cusp at the origin, touching the axis of x. 

By eliminating c obtain the differential equation of the family in 
the form 

ip 2 x 2 (x - 1 ) - ipxy (ix - 3y) + ( 1 6a; - 9) y 2 = 0. 

Show that both discriminants take the form x 3 y 2 =0, but that x=0 
is not a solution, while y =0 is a particular integral as well as an envelope. 

[This example shows that our theory does not apply without modi- 
fication to families of curves with a cusp at a fixed point.] 

(10) Show that the complete primitive of 

represents the family of equal lemniscates of Bernoulli 

r 2 = a 2 cos 2(0- a), 
inscribed in the circle r = a, which is the singular solution, with the 
point r = as a node-locus. 



80 DIFFERENTIAL EQUATIONS 

(11) Obtain and interpret the complete primitive and singular 
solution of /dr\ 2 



&) " 



\d6J 

(12) Show that r = cd-c 2 is the complete primitive and 4r = # 2 the 
singular solution of dr /g r \2 

r ' e Te-\Te)- 

Verify that the singular solution touches the complete primitive at 
the point (c 2 , 2c), the common tangent there making an angle tan -1 c 
vith the radius vector. 



CHAPTER VII 

MISCELLANEOUS METHODS FOR EQUATIONS OF THE 
SECOND AND HIGHER ORDERS 

68. In this chapter we shall be concerned chiefly with the 
reduction of equations of the second order to those of the first 
order. We shall show that the order can always be so reduced if 
the equation 

(i) does not contain y explicitly ; 

or (ii) does not contain x explicitly ; 

or (iii) is homogeneous. 

A special form of equation, of some importance in Dynamics, 
may be reduced by using an integrating factor. 

The remainder of the chapter will be devoted to the linear 
equation, excluding the simple case, already fully discussed in 
Chapter III., where the coefficients are merely constants. It will 
be found that the linear equation of the second order can be reduced 
to one of the first order if 

(i) the operator can be factorised, 
or (ii) any one integral belonging to the complementary function 
is known. 

If the complete complementary function is known, the equation 
may be solved by the method of Variation of Parameters. This 
elegant method (due to Lagrange) is applicable to linear equations 
of any order. 

Further information on linear equations, such as the condition 
for exact equations, the normal form, the invariantive condition of 
equivalence, and the Schwarzian derivative, will be found in the 
form of problems among the miscellaneous examples at the end 
of the chapter, with hints sufficient to enable the student to work 
them out for himself. 

P.D.E. 81 K 



82 DIFFERENTIAL EQUATIONS 

We shall use suffixes to denote differentiations with respect to 

dhi 
x, e.g. y 2 for tt|, but when the independent variable is any other 

than x the differential coefficients will be written in full. 

69. y absent. If y does not occur explicitly in an equation of 

the second order, write p for y x and J- for y 2 . 

We obtain an equation containing only J-, p, and x, and so of 
the first order. * 

Consider, for example, xy 2 +y 1 = ix. 

This transforms into Xj-+p=4:X, 

which can be integrated at once 

xp = 2x 2 + a, 

i.e. p = 2x + -. 
x 

By integrating, y = x 2 + a log x + b, 

where a and 6 are arbitrary constants. 

This method may be used to reduce an equation of the n th order 
not containing y explicitly to one of the (n - l) th . 

70. x absent. If x is the absent letter, we may still write p for 

. , , dp . dp dy dp dp ^ 

y v but for y 2 we now write V -f y , since V dy = TxWy =^ -y«- The 

procedure reduces an equation of the second order without x to one 
of the first order in the variables p and y. 
For example, W^Vx 

transforms into «/])-r= p 2 , 

from which the student will easily obtain 

p=by and y = ae hx . 

Examples for solution. 

(1) 2/ 2 cos 2 x = l. (2) yyz+y^-yv I (3) yy 2 + l=yi*. 

(4) Reduce to the previous example, and hence solve 

(5) xy 3 + y 2 = 12x. (6) 2/ n - 2 3/»-i = gX - 
(7) Integrate and interpret geometrically 

(i+yi 2 )* _ 7g 



EQUATIONS OF SECOND AND HIGHER ORDERS 83 

(8) The radius of curvature of a certain curve is equal to the length 
of the normal between the curve and the axis of x. Prove that the 
curve is a catenary or a circle, according as it is convex or concave to 
the axis of x. 

(9) Find and solve the differential equation of the curve the length 
of whose arc, measured from a fixed point A to a variable point P, is 
proportional to the tangent of the angle between the tangent at P and 
the axis of x. 

*71. Homogeneous equations. If x and y are regarded as of 
dimension 1, 

y x is of dimension 0, 
y 2 is of dimension - 1, 

y z is of dimension - 2, 
and so on. 

We define a homogeneous equation as one in which all the terms 
are of the same dimensions. We have already in Chap. II. dealt 
with homogeneous equations of the first order and degree, and in 
Chap. III. with the homogeneous linear equation 

x n y n + Ax n -iy n _ x + Bx n ~ 2 y n _ 2 + ...+ Exy x + Ky=0 
(where A, B, ... B, K are merely constants), for which we used the 
substitution x = (? or t =log x. 

Let us make the same substitution in the homogeneous equation 
xijy 2 +xy 1 2 =3yy 1 (1) 

Now v Ji%LjL*y 

Ul dx dt xdt' 

^ dx x 2 dt x dx dt 

_ 1 dy 1 dt d 2 y 
x 2 dt x dx dt 2 

l^dy \_d?y 

x 2 dt x 2 dt 2 ' 
Substituting in (1) and multiplying by x, we get 

y \dt 2 dt) + \dt) ~ 6y dt' 

d 2 y fdii\ 2 . du 

This is an equation, with t absent, similar to those in the last 
article with x absent. 






Arts. 71-73 may be omitted on a first reading. 



84 DIFFERENTIAL EQUATIONS 

By putting -tt = ?> the student will easily obtain 
yq=2(y 2 +b), 
giving t+c = l\og(y 2 +b). 

Hence y 2 +b=e^ t+c ^ 

= ax 4 , replacing e 4c by another arbitrary constant a. 

72. The example of Art. 71 came out easily because it had no 
superfluous a;'s left after associating x 2 with y 2 and x with y x . In 
fact, it could have been written 

y(*ty«) +(«&)" -3y(a^i). 

But (^+^ 2 )(2/-^i)+a;V«/ 2 =0 (2) 

cannot be so written. To reduce this to a form similar to that of 
the last example, put y=vx, a substitution used for homogeneous 
equations in Chap. II. 

(2) becomes 

(x 2 + x 2 v 2 ) (vx - v x x 2 - vx) + x i v 2 (xv 2 + 2uj) =0, 
i.e. -(1 + v 2 )v 1 +v 2 (xv 2 +2v 1 )=0, a. 

which may be written v 2 x 2 v 2 = (l -v 2 )xv 1 (3) 

"We now proceed as before and put x = e\ giving 

dv 

dH dv 
dfi'dt' 

(3) becomes ^--^=(1-^-, 

AH dv ,.v 

l - e ' V cU 2 =di> (4) 

an equation with t absent. 

, , dv d 2 v dq 

As before, put j-j, &-q& 

(4) becomes v 2 q^=q, 

i.e. t q = I (unless q = 0, giving y = ex), 
dv v i 



and x 2 v 2 =— 



dv = = 1_1 

dt V a v 



,, av dv / a 2 \, 

dt = = (o+- av, 

v - a \ v-a/ 



t=av + a 2 log (v-a) + b, 
and finally log x = ay/x + a 2 log (y - ax) - a 2 log x + b. 



EQUATIONS OF SECOND AND HIGHER ORDERS 85 

73. By proceeding as in the last article, we can reduce any 
homogeneous equation of the second order. 

Any such equation can be brought to the form 

ftyfayi>m/*)-o- 

For example, the equation of Art. 71 when divided by x becomes 
while that of Art. 72 divided by x 3 becomes 

(i*5)G-*)+©»-* 

The substitutions y =vx and x =e* transform 

/ (y/ x > Vx> x Vi) = to / (v, xv x + v, x 2 v 2 + 2xvj) = 0, 

j ,r /./ dv d 2 v dv\ n 

and then to /^, _ +v , _. + _j =0 , 

an equation with t absent, and therefore reducible to the first order. 

Examples for solution. 

(1) x 2 y 2 -xy 1 + y = 0. (2) x 2 y 2 -xy 1 + 5y = 0. 

(3) 2x 2 yy 2 + y 2 = x 2 y 1 2 . 

(i) Make homogeneous by the substitution y = z 2 , and hence solve 

2x 2 yy 2 + iy 2 = x 2 y x 2 + 2xyy v 

74. An equation occurring in Dynamics. The form y 2 =f{y) 
occurs frequently in Dynamics, especially in problems on motion 
under a force directed to a fixed point and of magnitude depending 
solely on the distance from that fixed point. 

Multiply each side of the equation by 2y v "We get 
2y 1 y 2 = 2f(y)y v 

Integrating, y 2 = 2 J / (y) £dx=2]f (y) dy. 

This is really the equation of energy. 

d 2 x 
dt 2 



o or 
Applying the method to -, z - = - p 2 x, (the equation of simple 



harmonic motion), we get 

z dtdt 2 ~ = ~ Zpx dt 

Integrating with respect to t, 

'dx , 2 



(dx 2 

( j- ) = -f 2 x 2 + const. =p 2 (a 2 -x 2 ), say 






86 DIFFERENTIAL EQUATIONS 



Hence 



<ti = l 1_ 

dx p \/(a 2 -x 2 )' 

1 . , x 
t = - sin -1 - + const., 
p a 

x = a sin (pt+e). 

Examples for solution. 

(1) y 2 = y 3 -y, given that y x =0 when y = \. 

(2) y 2 = e 2y , given that y = and y x = \ when x=0. 

(3) y 2 = sec 2 y tan y, given that ?/=0 and y x = \ when x = 0. " 

(l or nd (Lt 

(4) tt— - ^Tj given that x = A and -r=0 when < = 0. 
<w 2 x 2 at 

[h - x is the distance fallen from rest under gravity varying inversely 
as the square of the distance x from the centre of the earth, neglecting 
air resistance, etc.] 

... d 2 u P . . 

(5) ffli + u = fiw> m the two cases 

(i)P = Hu 2 ; (\\)P = ijm*; 

given that = -771 = when u — ~, where /x, h, and c are constants. 
(lu c 

[These give the path described by a particle attracted to a fixed 
point with a force varying inversely as the square and cube respectively 
of the distance r. u is the reciprocal of r, 6 has its ordinary meaning 
in polar co-ordinates, /u. is the acceleration at unit distance, and h is 
twice the areal velocity. ] 

)\ 75. Factorisation of the operator. The linear equation. 
(x + 2)y 2 - (2x+5)y 1 + 2y = (x + l)e x 
may be written as 

{(x + 2)D 2 - (2x +5)2) + 2}y = (x + l)e x , 

where D stands for -p, as in Chapter III. 

Now the operator in this particular example can be factorised, 
giving {(x+2)D-l}(D-2)y = (x + l)e x . 

Put (D-2)y=v. 

Then {(x+2)D -l}0-(x + l)&. 

This is a linear equation of the first order. Solving as in Art. 20, 
we get v = c{x+2)+e x , 

i.e. (D-2)y = c(x+2)+e x , 
another linear equation, giving finally 

y = a(2x + 5) + be 2x - e x , replacing - \c by a. 



EQUATIONS OF SECOND AND HIGHER ORDERS 87 

Of course it is only in special cases that the operator can be 
factorised. It is important to notice that these factors must be 
written in the right order, as they are not commutative. Thus, on 
reversing the order in this example, we get 

(D-2){(x + 2)D-l}y = {(x+2)D 2 -(2x + 4)D + 2}y. 

Examples for solution. 

(1) (x + l)y 2 + (x-l) yi -2y=0. (2) xy 2 + (x-\) yi -y = 0. 

(3) xy 2 + (x-l)y 1 -y = x 2 . 

(4) xy 2 + (x 2 + l)y l + 2xy = 2x, given that y = 2 and yx = when 
x=0. 

(5) (x 2 - 1) y 2 - (4a; 2 - 3x - 5) y x + (4z 2 - 6x - 5) y = e 2x , given that y = 1 
and «/! = 2 when £ = 0. 

76. One integral belonging to the complementary function * known. 

When one integral of the equation 

y 2 + Py, + Qy=0 (1) 

is known, say y = z, then the more general equation of the second 

order y 2 + Py 1 + Qy=R, (2) 

where P, Q, R are functions of x, can be reduced to one of the first 
order by the substitution « _ VZt 

Differentiating, y 1 = v ± z + vz 1} 

y 2 = v 2 z+2v 1 z 1 +vz 2 . 
Hence (2) becomes 

t>gz + v x (2z 1 +Pz)+v(z 2 +Pz 1 +Qz)=0, 

i.e. z^+v^z.+Pz)^, (3) 

since by hypothesis z 2 + Pz 1 +Qz=0. 

(3) is a linear equation of the first order in v v 

Similarly a linear equation of the n th order can be reduced to 
one of the (n - 1 ) th if one integral belonging to the complementary 
function is known. 

77. Example. 

Consider again the equation 

(x + 2)y 2 -(2x + 5)y 1 + 2y = (x + l)e x (4) 

*The proof of Art. 29 that the general solution of a linear differential equation is 
the sum of a Particular Integral and the Complementary Function holds good when 
the coefficients are functions of x as well as in the case when they arc constants. 



88 DIFFERENTIAL EQUATIONS 

If we notice that y = e 2x makes the left-hand side of the equation 
zero, we can put y = ve 2x 

giving y 1 = (v 1 + 2v)e 2x , 

and y 2 = (v 2 + 4v, + 4v) e 2x . 

Substitution in (4) gives 

(x + 2)v 2 e 2x + {Mx + 2)-(2x + 5)}v 1 e 2x 

+{i(x + 2)-2(2x + 5)+2}ve 2x = (x + l)e x , 

dv 
i.e. (x + 2)j 1 + {2x + 3) v 1 = {x + l)e~ x 

Solving this in the usual way (by finding the integrating factor) 
we obtain Vi = e~ x + c(x + 2) e~ 2x . 

Integrating, v= - e~ x - \c{2x + 5) e~ 2x + b, 

whence y = ve 2x = - e x - ±c(2x + 5) + be 2x . 

Examples for solution. 

(1) Show that y 2 + Pyx+Qy=0 is satisfied by y = e x if 1+P + Q=0, 
and by y = x if P + Qx = 0. 

(2) x 2 y 2 + xy 1 -y = 8x 3 . 

(3) x 2 y 2 -(x 2 + 2x)y 1 + (x + 2)y = x s e x . 

(4) xy 2 -2 (x + l)y 1 + (x + 2) y = (x-2)e 2x . 

(5) x 2 y 2 + xy 1 -9y = 0, given that y = x* is a solution. 

(6) xy 2 (x cos x-2 sin x) +(x 2 + 2)y x sin x-2y (x sin z + cos x) =0, 
given that y = x 2 is a solution. 

78. Variation of Parameters. We shall now explain an elegant 
but somewhat artificial method for finding the complete primitive 
of a linear equation whose complementary function is known. 

Let us illustrate the method by applying it to the example 
already solved in two different ways, namely, 

(x+2)y 2 -(2x+5)y 1+ 2y = (x + l)e*, ...(1) 

of which the complementary function is y = a(2x +5) +be 2x . 

Assume that y = (2x+5)A +e 2x B, (2) 

where A and B are functions of x. 

This assumption is similar to, but more symmetrical than, that 
of Art. 77, viz. : y^^x 

Differentiating (2), 

y 1 =(2x +5)A t +e 2x B l +2A +2e 2x B (3) 

Now so far the two functions (or parameters) A and B are only 
connected by the single equation (1). We can make them satisfy 
the additional equation 

(2x+5)A 1 +e 2x B l =0 (4) 



EQUATIONS OF SECOND AND HIGHER ORDERS 89 

(3) will then reduce to 

y 1 =2A+2e 2x B (5) 

Differentiating (5), 

y 2 =±e Zx B + 2A 1 +2e 2x B l (6) 

Substitute these values of y, y v and y 2 from equations (2), (5), 
and (6) respectively in (1). The co-factors of A and B come to 
zero, leaving 

2{x+2) A l +2(x + 2)e 2x B 1 =(x + l)e x (7) 

(4) and (7) are two simultaneous equations which we can solve 
for A x and B v giving 

4l B i (x + l)e x __ (x + l)er x 

e 2x~ -(2x +b)~2e 2x (x +2)(1 -2x-5)~ 4(z+2) 2 ' 
w . (x + l)e x _ e x j 1 1 1 • 

Mence ^i- 4(* + 2)«"~Tls+2 (^T2) 2 /' 

€ x 

and. by integration, A = - -r-. ^ + a, where a is a constant. 

J & 4 (a; +2) 

Similarly, 

R = (^x+5){x + l)e- x er*_ f _ _1_ _1 I 

x ~ , 4(z+2) 2 "4 I x+2 (ic+2) 2 /' 

and B = ~{-\-2\+b. 

4 \x+2 J 

Substituting in (2), 

y=v*H -i(£2) +a } + ?{^2- 2 l +be " 

= a(2x+5)+be 2x -e x . 

79. Applying these processes to the general linear equation of 

the second order, y 2 + Py 1 + Qy = R, (1) 

of which the complementary function au+bv is supposed known,- 
a and b being arbitrary constants and u and v known functions of x r 

we assume that y=uA+vB, (2) 

giving y 1 =u l A+v 1 B, (3) 

provided that uA 1 +vB 1 =0 (4) 

Differentiating (3), 

y 2 = u 2 A + v 2 B + u 1 A 1 +v l B 1 (5) 

Substitute for y 2 , y l and y in (1). 

The terms involving A will be A(u 2 + Pu 1 + Qu), i.e. zero, as by 
hypothesis, Wg + p % + q u = . 

Similarly the terms involving B vanish, and (1) reduces to 

u x A x +v 1 B l =R (6) 



90 DIFFERENTIAL EQUATIONS 
Solving (4) and (6), — * = A = ^_ 

V -U VU X -UV X * 

We then get A and B by integration, say 

A=f(x)+a, 
B=F(x)+b, 
where / (x) and F (x) are known functions of x, and a and 6 are 
arbitrary constants. 

Substituting in (2), we get finally 

y = uf(x) + vF (x) + au + bv. 

* 80. This method can be extended to linear equations of any 
order. For that of the third order, 

y 3 + Py 2 + Qy 1 + Ry=S, (1) 

of which the complementary function y = au +bv+cw is supposed 
known, the student will easily obtain the equations 

y = uA+vB+wC, (2) 

y 1 =u 1 A +v 1 B+w 1 C, (3) 

provided that 0=uA 1 +vB 1 +ivC 1 ; (4) 

hence y 2 = u 2 A+v 2 B +w 2 C, (5) 

provided that 0=u 1 A 1 +v 1 B 1 +w 1 C 1 ; (6) 

then y z - u 3 A + v 3 B + w 3 C 

+ u 2 A 1 +v 2 B 1 +w 2 C 1 ; (7) 

by substitution in(l), S=v 2 A 1 +v 2 B 1 +w 2 C 1 (8) 

A ly B ly and C x are then found from the three equations (4), (6) 
and (8). 

Examples for solution. 

(1) 2/2 + V = cosec x. (2) y 2 + iy = 4 tan 2z. 

< 3 ) y*-y=rh' 

(4) x 2 y 2 + xy x - y = x 2 e x , given the complementary function ax + bx* 1 . 

(5) y 3 -6ij 2 + n yi -6y = e 2 *. 

81. Comparison of the different methods for solving linear equations. 
If it is required to solve a linear equation of the second order and 
no special method is indicated, it is generally best to try to guess 
a particular integral belonging to the complementary function and 
proceed as in Art. 76. This method may be used to reduce a linear 
equation of the n th order to one of the (n - l) th . 

* To be omitted on a first reading. 



EQUATIONS OF SECOND AND HIGHER ORDERS 91 

The method of factorisation of the operator gives a neat solution 
in a few cases, but these are usually examples specially constructed 
for this purpose. In general the operator cannot be factorised. 

The method of variation of parameters is inferior in practical 
value to that of Art. 76, as it requires a complete knowledge of the 
complementary function instead of only one part of it. Moreover, 
if applied to equations of the third or higher order, it requires too 
much labour to solve the simultaneous equations for A x> B v C x , etc., 
and to perform the integrations. 

MISCELLANEOUS EXAMPLES ON CHAPTER VII. 

(1) 2/^2 ~2/i 2 + 2/i = 0. <2) xy 2 + xy 1 2 -y 1 = 0. 

(3) y»*r*y*-v ( 4 ) y„ + «/„- 2 = 8cos3z. 

(5) (x 2 log x - x 2 )y 2 - xy t + y = 0. 

(6) (x 2 + 2x -l)y 2 - (3x 2 + 8x -l)y 1 + (2x 2 + 6x)y = 0. 

(7) Verify that cos nx and sin nx are integrating factors of 

y 2 + n 2 y=f{x). 
Hence obtain two first integrals of 

y 2 + n 2 y = sec nx, 
and by elimination of y x deduce the complete primitive. 

(8) Show that the linear equation 

Ay + By 1 + Cy 2 + ...+Sy n = T, 

where A, B, C, ... T are functions of x, is exact, i.e. derivable imme- 
diately by differentiation from an equation of the next lower order, if 
the successive differential coefficients of A, B, C, ... satisfy the relation 

A-B 1 + C 2 -...+(-l) n S n = 0. 

[N.B. — By successive integration by parts, 

]Sy n dx = Sy n ^ 1 -S 1 y n _ 2 + S 2 y n _ 3 + ...+(-l) n - 1 S n _ 1 y + }(-l) n S n ydx.] 

Verify that this condition is satisfied by the following equation, and 
hence solve it : 

(2x 2 + 3x) y 2 + (6x + 3)y 1 + 2y = (x + l)e x . 

(9) Verify that the following non-linear equations are exact, and 
solve them: (i) yy 2 + yi 2 = 0. 

(ii) xyy 2 + xy 1 2 + tjy 1 = 0. 

(10) Show that the substitution y = ve •> IX transforms 
y 2 + Py 1 +Qy = R, 
where P, Q, and R are functions of x, into the Normal Form 

v 2 + Iv = S, 



92 DIFFERENTIAL EQUATIONS 

where I=Q-\P X -\P\ 

and S = Re^ Pdx . 

Put into its Normal Form, and hence solve 

y 2 -ixy 1 + (4:X 2 -l)y= - 3e x * sin 2x. 

(11) Show that if the two equations 

and z 2 + pz 1 + qz = 

reduce to the same Normal Form, they may be transformed into 

each other by the relation 

\[pdx \\pdx 

i.e. the condition of equivalence is that the Invariant I should be the 
same. 

(12) Show that the equations 

x 2 y 2 + 2(x z -x)y 1 + (l-2x 2 )y=0 
and x 2 z 2 + 2(x 3 + x)z l -{l-2x 2 )z=0 

have the same invariant, and find the relation that transforms one into 
the other. Verify by actually carrying out this transformation. 

(13) If u and su are any two solutions of 

v 2 + lv = 0, (1) 

prove that ^ = _2^1, (2) 

s 1 u 

and hence that ^-^(- 2 Y = 2/ (3) 

From (2) show that if s is any solution of (3), s^ and ss^ are 
solutions of (1). 

[The function of the differential coefficients of s on the left-hand 
side of (3) is called the Schivarzian Derivative (after H. A. Schwarz of 
Berlin) and written {s, x}. It is of importance in the theory of the 
Hypergeometric Series.] 

(14) Calculate the Invariant / of the equation 

x 2 y 2 - (x 2 + 2x)y x + (x + 2)y=Q. 
Taking s as the quotient of the two solutions xe x and x, verify that 

{s,x} = 21, 

and that s x and ssf* are solutions of the Normal Form of the original 
equation. 

(15) If u and v are two solutions of 

y 2 +Pyi+Qy=o, 

prove that uv 2 - vu 2 + P{uv 1 - vuj) = 0, 

and hence that uv 1 -vu 1 — ae * . 

Verify this for the equation of the last example. 



MISCELLANEOUS EXAMPLES 93 

(16) Show that yy x = const, is a first integral of the equation formed 
by omitting the last term of 

By putting yy x = C, where C is now a function of x (in fact, varying 
the parameter C), show that if y is a solution of the full equation, then 

c[=-y 2 , 
and hence C 2 = const. - \y*, 

giving finally y 2 = a sin {x\J1 + b). 

[This method applies to any equation of the form 

*■+*■/(*) +*(y)-o.] 

(17) Solve the following equations by changing the independent 
variable : 

d 2 ti Q'V 

(ii) (l+*«)»g + 2»(l+^+4y-a 

(18) Transform the differential equation 

j~2 cos x + ~j~ sm x ~ %y cos3 X = ^ c °s 5 £ 

into one having z as independent variable, where 

z = sin x, 
and solve the equation. [London.] 

(19) Show that if z satisfies 

^ + P-=0, 
dx 2 dx 

by changing the independent variable from x to z, we shall transform 

into d4 + Sy=T ' 

Hence solve + (l -^)^ + ix 2 ye- 2x = 4:(x 2 + x 3 )e- Zx . 



CHAPTER VIII 

NUMERICAL APPROXIMATIONS TO THE SOLUTION OF 
DIFFERENTIAL EQUATIONS 

82. The student will have noticed that the methods given in the 
preceding chapters for obtaining solutions in finite form only apply 
to certain special types of differential equations. If an equation 
does not belong to one of these special types, we have to use approxi- 
mate methods. The graphical method of Dr. Brodetsky, given in 
Chapter L, gives a good general idea of the nature of the solution r 
but it cannot be relied upon for numerical values. 

In this chapter we shall first give Picard's * method for getting 
successive algebraic approximations. By putting numbers in these, 
we generally get excellent numerical results. Unfortunately the 
method can only be applied to a limited class of equations, in which 
the successive integrations can be easily performed. 

The second method, which is entirely numerical and of much 
more general application, is due to Runge.f With proper pre- 
cautions it gives good results in most cases, although occasionally 
it may involve a very large amount of arithmetical calculation. "We 
shall treat several examples by both methods to enable their merits 
to be compared. 

Variations of Runge's method have been given by Heun, Kutta, 
and the present writer. 

83. Picard's method of integrating successive approximations. The 

differential equation d v 

* E. Picard, Professor at the University of Paris, is one of the most distinguished 
mathematicians of to-day. He is well known for his researches on the Theory of 
Functions, and his Cours (Tanalyse is a standard text-book. 

t C. Runge, Professor at the University of Gottingen, is an authority on 
graphical methods. 

94 



NUMERICAL APPROXIMATIONS 9fr 

where y =6 when x-a, can be written 

*/ = & + [ f(x,y)dx. 

For a first approximation we replace the y inf(x, y) by b ; for 
a second we replace it by the first approximation, for a third by the 
second, and so on. 

du 
Ex. (i). J?- = x + y 2 , where y=0 when x=0. 

Here y = I (x + y 2 ) dx. 

Jo 

First approximation. Put ?/=0 in x + y 2 , giving 

y=\ xdx = \x 2 . 
Jo 

Second approximation. Put ?/ = \x 2 va.x + y 2 , giving 

r 

y = (» + iz 4 ) (far = \x 2 + z\x 5 . 
Jo 

Third approximation. Put i/ = \x 2 + -^V^ 5 in cc + «/ 2 , giving 

r 

1/ = (a? + \& + -itx 1 + T hvx 10 ) dx 
Jo 

— i r 2 i 1 r 5 i 1 y 8 i l yll 

— 2 X ^W*' + TTRr ,c + TTTTD" X > 

and so on indefinitely. 

where t/ = l and z = £ when cc=0. 

Here ?/ = l + l 2^ and z = i+l x 3 (y + z)dx. 

Jo Jo 

First approximation. 

y=*l + I -^cfa; = l+£x, 

Jo" 

z = \ + [V(l +|) cfc^ + fz 4 . 
Jo 

Second approximation. 

y = l + [ (% + %x i )dx = l+±x + T s 1T x 5 , 



■96 DIFFERENTIAL EQUATIONS 

Third approximation. 

y - 1 + f (£ + f x* + -As 5 + frx*) dx 
Jo 



- 1 + \x + ^z 5 + isVz 6 + T fox 9 , 

z =4 + r a»(4 + \x + f z 4 + ^r 5 + -fox 8 ) dx 
Jo 

and so on. 

Ex. (iii). j^ = x 3 (^- + y), where */ = l and -j-=\ when x=0. 

By putting j-~z, we reduce this to Ex. (ii). 

It may be remarked that Picard's method converts the differential 
equation into an equation involving integrals, which is called an Integral 
Equation. 

Examples for solution. 

Find the third approximation in the following cases. For examples 
(1) and (2) obtain also the exact solution by the usual methods. 

\/ (1) ~ = 2y - 2x 2 - 3, where y = 2 when x = 0. 

CLOC 

(2) -~ = 2--, where v = 2 when x = l. 
dx x 

ft-"*-. 

where y = 2 and z=0 when x = 0. 

(dy 



(4) 



dx~ Z ' 

dz a . 

- 1 -=x 2 z + x*y, 
dx 9 



where y = 5 and 2 = 1 when x=0. 

(5) y = x zJ- + tfy where y = 5 and / = 1 when x=0. 
v ' dx 2 dx * * dx 

84. Determination of numerical values from these approximations. 

Suppose that in Ex. (i) of the last article we desire the value of y, 
correct to seven places of decimals, when x =0-3. 

Substituting x =0-3, we get \ (0-3) 2 =0-045 from the first approxi- 
mation. 

The second adds ^(0-3) 5 =0-0001215, 
while the third adds ^(O^) 8 + tt Vtf(0-3) u =0-00000041 ... . 



NUMERICAL APPROXIMATIONS 



97 



Noticing the rapid way in which these successive increments 
decrease, we conclude that the next one will not affect the first 
seven decimal places, so the required value is 0-0451219... . 

Of course for larger values of x we should have to take more 
than three approximations to get the result to the required degree 
of accuracy. 

We shall prove in Chap. X. that under certain conditions the 
approximations obtained really do tend to a limit, and that this limit 
gives the solution. This is called an Existence Theorem. 

Example for solution. 

(i) Show that in Ex. (ii) of Art. 83, x = 0-5 gives y = 1-252... and 
2=0-526... , while z=0-2 gives y = 1-100025... and 2 = 0-500632... . 

85. Numerical approximation direct from the differential equation. 

The method of integrating successive approximations breaks down 
if, as is often the case, the integrations are impracticable. But 
there are other methods which can always be applied. Consider 
the problem geometrically. The differential equation 

dy 



dx 



=/fo y) 



determines a family of curves (the " characteristics ") which do not 
intersect each other and of which one passes through every point 




Fig. 23. 

in the plane.* Given a point P (a, b), we know that the gradient 
of the characteristic through P is f (a, b), and we want to determine 

* This is on the assumption that f(x, y) has a perfectly definite value for every 
point in the plane. If, however, f(x, y) becomes indeterminate for one or more 
points, these points are called singular points of the equation, and the behaviour 
of the characteristics near such points calls for special investigation. See Art. 10. 



98 DIFFERENTIAL EQUATIONS 

the y =NQ of any other point on the same characteristic, given that 
x = ON = a + h, say. A first approximation is given by taking the 
tangent PR instead of the characteristic PQ, i.e. taking 

y =2VX +LR=NL+PL tan /_RPL = b+hf{a, b) =b + hf , say. 

But unless h is very small indeed, the error RQ is far from 
negligible. 

A more reasonable approximation is to take the chord PQ as 
parallel to the tangent to the characteristic through S, the middle 
point of PR. 

Since <# is (a + \h, b + %hf ), this gives 

y =NL +LQ=NL + PL tan /_QPL =b+hf{a + \h, b +ffi ). 

This simple formula gives good results in some cases, as will be 
seen from the following examples : 

Ex. (i) -^- = x + y 2 ; given that y=0 when x = 0, required y when 
a; = 0-3. dx 

Here a = 6 = 0, A = 0-3, f(x,y)=x + y*. 

Therefore 

/o=/(a,6)=0, a + £A = 0-15, 6 + P/ = 0, 

giving b + hf{a + \h, 6 + ^/ )=0 + 0-3 x/(0-15, 0) =0-045. 

The value found in Art. 84 was 0-0451219... , so the error is 
0-00012... , about J per cent. 

Ex. (ii). -j- = 2 - - ; given that y — 2 when x = l, find y when x = 1 -2. 
Here o-l, 6 = 2, A=0-2, / =2-f=0. 

Therefore b + hf(a + $h, 6 + ^/ ) = 2 + 0-2 x/(M, 2) 

= 2+0-2 x (2--^-) =2-036.... 
Now the differential equation is easily integrable, giving y = x + -, 

so when a? = 1-2 the value of y is 2-033... . The error is 0-003... , which 
is rather large compared with the increment of y, namely 0-036... . 

Ex. (iii). -g = «=/(&, y, z), say, 

-^ = x 3 (y + z)~g(x, y, 2), say; 

given that y = \ and 2 = 0-5 when x = 0, find y and 2 when x = 0-5. 
Here a=0, 6 = 1, c (the initial value of 2) =0-5, A=0-5. 
Hence / =/(0, 1, 0-5) =0-5 ; g =g<0, 1, 0-5)=0. 



NUMERICAL APPROXIMATIONS 99 

By an obvious extension of the method for two variables, we take 

= b + hf{a + \h, b + \hf , c + ^ ) = l+0-5x/(0-25, 1-125, 0-5) = l-2500, 

and z = c + hg(a + $h, b + Wo> c + ihg Q ) 

=0-5 +0-5x0(0-25, 1-125, 0-5) =0-5127. 

The accurate values, found as in Art. 84, are 

y = 1-252... and 2=0-526.... 

Thus we have obtained a fairly good result for y, but a verv bad 
one for z. J 

k The il un J cer * aiut y ab °ut the degree of accuracy of the result deprives 
the method of most of its value. However, it forms an introduction to 
articb° re e meth ° d 0f Run S e > t0 be explained in the next 

Examples for solution. 

.-. dy i 

{) dx = {x ~ y) ~ 1; ^ven that y = 4 when x = 2-3, obtain the value 
y = 4-122 when z = 2-7. [Runge's method gives 4-118.] 

in\ dy 1 $ 

{) di^lO™ - 1+lo &( x + y)}'> given that y = 2 when x=-l, obtain 
;he value y = 2-194 when a-1. [Runge's method gives 2-192.] 

ti\ dy y • 

{) ti '< x' g lventhat 2/ = 2whenz = l, obtain the valuey = 2-076 

X ion 2 :. .^ show that y =l x2+ i> so that when - 1 * V ™ 

86. Runge's method. Suppose that the function of y defined * by 

~dx ~$ ^' $' y =b when x = a > 

3 denoted by y=F(x). 

If this can be expanded by Taylor's theorem, 

F(a + h) =F{a) + hF{a) +^F"(a) +~F'"(a) + ... . 

Now F\x)=%=f{x,y)=f,^y. 

We shall now take the total differential coefficient with respect 
> x (that is, taking the y in/ to vary in consequence of the variation 
i x). ^ Let us denote partial differential coefficients by 

p-f?, ,-§?, rJ-f s-^f t- d2 f- 
to' * dy' dx*' S ~dxdy' t dtf> 

id their values when x = a and y =b by Po , q Q , etc. 

io^IinodZ.Tt' T hlCh th « diffe ^«al equation and the initial con- 
■«♦ «?lu 7 i * 1 • f functl0n are discussed in Chap. X. The graphical treat 
nt of the last article assumes that these conditions are satisfied graptU ° al treat " 



100 DIFFERENTIAL EQUATIONS 

Similarly, *»<(*) = (| + | J) (,+,,) 

= r+^+/s+/(s+ ? 2+/|F). 
Thus 
JP(a+A)-.F(a) 

= fc/ + JA 8 (p +/o?o) +#"(*• +2/>o +/o% +M> +/o?o 2 ) + ••• • (1) 
The first term represents the first approximation mentioned and 
rejected in Art. 85. 

The second approximation of Art. 85, i.e. 

y-b=hf{a+\h,b + \hf Q ) = k 1} say, 
may now be expanded and compared with (1). 

Now, by Taylor's theorem for two independent variables, 
f(a+lh,b+±hf ) x 

=/o + 1% + Wtfo + 2! (^ 2/ "o + hh 2 Uo + i Wo) + • • • > 

giving A; 1 =A/ +^ 2 (^o+/o?o)+^ 3 K+2/oSo4o) + ( 2 ) 

It is obvious that Jc x is at fault in the coefficient of h 3 . 

Our next step is suggested by the usual methods * for the 

numerical integration of the simpler differential equation 

!=/(*>■ ■ 

Our second approximation in this case reduces to the Trapezoidal 
Rule y-b=hf{a+\h). 

Now the next approximation discussed is generally Simpson's 
Rule, which may be written 

y-&-i*{/(a)+4/(a+#)+/(a + *)}. 
If we expand the corresponding formula in two variables, namely 
\Hh + tfiP+\K b+lhf^+fia+h, b + hf )}, 
we easily obtain 

¥o + ¥i 2 (Po+foqo)+hW(ro + 2foSo + t ) + ---, (3) 

which is a better approximation than k v but even now has not the 
coefficient of h 3 quite in agreement with (1). 

To obtain the extra terms in h 3 , Runge f replaces 
hf(a+h,b+hf ) 

* See the text- books on Calculus by Gibson or Lamb, 
f Mathematische Annalen, Vol. XL VI. pp. 167-178. 



NUMERICAL APPROXIMATIONS 101 

by V" = hf(a +h,b+ V), where k" = hf(a + h,b+ hf ). The modified 
formula maybe briefly written ^{k' +ik l +k'"}, where k' =hf , or 
f k x + ^k 2 = k x + 1 (k 2 - kj), where k 2 = \(k' +k'"). 

v The student will easily verify that the expansion of Runge's 
formula agrees with the right-hand side of (1) as far as the terms 
in h, h 2 , and h 3 are concerned. 

Of course this method will give bad results if the series (1) con- 
verges slowly. 

If y o > 1 numerically, we rewrite our equation 

i = Khr F(x ' y) - m ' 

and now F < 1 numerically, and we take y as the independent 
variable. 

87. Method of solving examples by Runge's rule. To avoid 
confusion, the calculations should be formed in some definite order, 
such as the following : 

Calculate successively k' =hf , 

k"=hf{a + h,b+k'), 
k"'=hf(a + h,b+k"), 
k^hfia + ih, 6+P'), 
k 2 = 2 \k +k ), 
and finally k = k 1 +^(k 2 -k 1 ). 

Moreover, as k x is itself an approximation to the value required, 
it is clear that if the difference between k and k v namely I (k 2 - k x ), 
is small compared with k ± and k, the error in k is likely to be even 
smaller. 

dy 
Ex. (i). j-=x + y 2 ; given that y=0 when x = 0, find y when x = 0-3. 

Here a = 0, 6 = 0, A = 0-3, f(x,y)=x + y 2 , / = 0; 
tf-Vo-0; 

k" = hf{a + h, & + A/)=0-3x/(0-3, 0)=0-3x0-3 =0-0900; 

k'" = hf(a + h, b + k") =0-3 x/(0-3, 0-09) =0-3 x (0-3 + 0-0081) =0-0924 ; 

k x -hf{a + \h, 6 + P') =0-3 x/(0-15,0) =0-3x0-15 =0-0450; 

k 2 = \(k' + k'") = \x 0-0924 =0-0462; 

and 

k = k x +£(& 2 -fc 1 ) =0-0450 + 0-0004 =0-0454. 

As the difference between k = 0-0454 and k x = 0-0450 is fairly small 
compared with either, it is highly probable that the error in k is less 



102 DIFFERENTIAL EQUATIONS 

than this difference 0-0004. That is to say, we conclude that the value 
( is 0-045, correct to the third place of decimals. 

We can test this conclusion by comparing the result obtained in 
Art. 84, viz. 0-0451219... . 

fjbti II — nt* 

Ex. (ii). j- ; given that y = \ when sc = 0, find y when x = l. 

(too y *t" x 

This is an example given in Eunge's original paper. Divide the 
range into three parts, to 0-2, 0-2 to 0-5, 0-5 to 1. We take a small 
increment for the first step because / (x, y) is largest at the beginning. 
First step. a = 0, 6 = 1, /* = 0-2, / = 1 ; 

k' = hf =0-200 

k" = hf(a + h, b + k') =0-2 xf (0-2, 1-2) =0-143 

k'" = hf{a + h, b + k")=0-2xf(0-2, 1-143) =0-140 

k x = hf(a + \h,b + \k')=0-Zxf{0-\, 1-1) =0-167 

k 2 = ^{k' + k'") = ^x 0-340 =0-170 

and k^^ + H^-kj) =0-167 +0-001 =0-168 

giving i/ = l-168 when z = 0-2. 

Second step. 

a = 0-2, 6 = 1-168, 7*=0-3, / =/(0-2, 1-168) =0-708. 
Proceeding as before we get i 1 = 0-170, & 2 = 0-173 and so £ = 0-171, 
giving y = 1-168 +0-171 = 1-339 when x = 0-5. 

Third step. a = 0-5, 6 = 1-339, 7j = 0-5. 

We find k t = k 2 = k = 0-160, giving i/ = l-499 when x = l. 
Considering the k and k v the error in this result should be less than 
0-001 on each of the first and second steps and negligible (to 3 decimal 
places) on the third, that is, less than 0-002 altogether. 

As a matter of fact, the true value of y is between 1 -498 and 1 -499, 
so the error is less than 0-001. This value of y is found by integrating 
the equation, leading to 

7T - 2 tan- 1 ^ = log e (x 2 + y 2 ). 

Examples for solution. 

Give numerical results to the following examples to as many places 
of decimals as are likely to be accurate : 

(1) y = 77;{y -1 +l°ge(3 + 2/)} ; given that y = 2 when x= -1, find 

y when x = l, taking h = 2 (as /is very small). , 

(2) Obtain a closer approximation to the preceding question by 
taking two steps. 

(3) ~ = (x 2 -y) -1 ; given that y = i when x = 2 -3, find y when 
x = 2 -7 (a) in one step, (6) in two steps. 






NUMERICAL APPROXIMATIONS 103 

dtf V 

(4) Show that if -^=2-- and y = 2 when as—1, then y = x + -. 

dx X X 

Hence find the errors in the result given by Runge's method, taking 
(a) ^=0-4, (b) h=0-2, (c) h=0-l (a single step in each case), and compare 
these errors with their estimated upper limits. 

(5) If E(h) is the error of the result of solving a differential equation 
of the first order by Runge's method, prove that 

Lt E W _ I 

h .+o E{nh) n*' 

Hence show that the error in a two-step solution should be about 
■§■ of that given by one step ; that is to say, we get the answer correct 
to an extra place of decimals (roughly) by doubling the number of steps. 

88. Extension * to simultaneous equations. The method is easily 
extended to simultaneous equations. As the proof is very similar 
to the work in Art. 86, though rather lengthy, we shall merely give 
an example. This example and those given for solution are taken, 
with slight modifications, from Runge's paper. 

Ex. £ = 2 2 -|=/(x, y, z), say, 

dz y . 

given that y= 0-2027 and 2 = 1-0202 when x = 0-2, find y and z when 
z = 0-4. 
Here 
o = 0-2, 6=0-2027, c = 1-0202, /„=/ (0-2, 0-2027, 1-0202) = 1-027, 
a = 0-2070, A = 0-2; 
k' = hf = 0-2x 1-027 =0-2054 

Z' = ^ = 0-2x 0-2070 =0-0414 

k" = hf(a + h, b + k', c + V) =0-2 x/(0-4, 0-4081, 1-0616) =0-2206 
l" = hg(a + h, b + k', c + l') =0-2 xo(0-4, 0-4081, 1-0616) =0-0894 
k'" = hf(a + h, b + k", c + O=0-2x/(0-4, 0-4233, 1-1096) =0-2322 
l'" = hg(a + h, b + k", c + l") =0-2 xo(0-4, 0-4233, 1-1096) =0-0934 
kj_ = hf(a + \h, b + p', c + \V) =0-2 x/(0-3, 0-3054, 1 -0409) =0-2128 
l^hfia + lh, b + \lc', c + \l') =0-2xo(0-3, 0-3054, 1-0409) =0-0641 
k 2 = $(k' + k'") =0-2188 

1 2 = \{V + V") =0-0674 

k = k 1 + ^(k 2 -k 1 )= 0-2128 + 0-0020 =0-2148 

l = ^ + l(l 2 -Ji) =0-0641 +0-0011 =0-0652 

giving y = 0-2027 +0-21 48 = 0-41 75 

and 2 = 1 -0202 + -0652 = 1 -0854, 

probably correct to the third place of decimals. 

*The rest of this chapter may be omitted on a first reading. 



104 DIFFERENTIAL EQUATIONS 

Examples for solution. 

(1) With the equation of Art. 88, show that if y =04175 and 
2 = 1-0854 when x =0-4, then y =0-6614 and 2 = 1-2145 (probably correct 
to the third place of decimals) when x = 0-6. 

... dw _ V(l-w 2 ) dr w . , 

(2) — =-2« + -^ '-; j-^—^z r x ; given that w =0-7500 

dz r dz v(l ~ w ) 

and r=0-6 when 2 = 1-2145, obtain the values w =0-5163 and r =0-7348 

when 2 (which is to be taken as the independent variable) = 1-3745. 

Show that the value of r is probably correct to four decimal places, but 

that the third place in the value of iv may be in error. 

(3) By putting w = cos <f> in the last example and t/ = sin <p, x = r in 
the example of Art. 88, obtain in each case the equations 

dz _ sin d> d<h 

^- = tanrf>; 22 = - + cos<b-r-, 

dr r r T dr 

which give the form of a drop of water resting on a horizontal plane. 

89. Methods* of Heun and Kutta. These methods are very 
similar to those of Runge, so we shall state them very briefly. The 

problem is : given that -~ =f(x, y) and y=b when x = a, to find 

the increment h of y when the increment of x is h. 
Heun calculates successively 

k'=hf(a,b), 
k"=hf(a+ih,b+lk'), 
Jc'"=hf(a + %h,b+ik"), 
and then takes l(k' +SJc'") as the approximate value of h. 
Kutta calculates successively, 
k'=hf(a,b), 
Jc"=hf{a+ih,b+ik'), 
k'"=hf(a+lh,b+k" -\k'), 
k""=hf(a+h,b+k"'-k"+k'), 
and then takes |(&' +3k" +3k'" +k"") as the approximate value 
oik. 

The approximations can be verified by expansion in a Taylor's 
series, as in Runge's case. 

Example for solution. 

n / 'U <u nr 

Given that -f-=- and y — \ when x=0, find the value of y (to 8 

significant figures) when x = 1-2 by the methods of Runge, Heun, and 
Kutta, and compare them with the accurate value 1-1678417. [From 
Kutta 's paper. ] 

* Zeitschrift fur Mathematik und Physik, Vols. 45 and 46. 



NUMERICAL APPROXIMATIONS 



105 



90. Another method, with limits for the error. The present writer 
has found * four formulae which give four numbers, between the 
greatest and least of which the required increment of y must lie. 
A new approximate formula can be derived from these. When 
applied to Runge's example, this new formula gives more accurate 
results than any previous method. 

The method is an extension of the following well-known results 
concerning definite integrals. 

91. Limits between which the value of a definite integral lies. Let 

F(x) be a function which, together with its first and second 
differential coefficients, is continuous (and therefore finite) between 
x=a and x=a + h. Let F"{x) be of constant sign in the interval. 
In the figure this sign is taken as positive, making the curve concave 
upwards. LP, MQ, NR are parallel to the axis of y, M is the 
middle point of LN, and SQT is the tangent at Q. OL=a, LN = h. 





3 










R 


s 


~-^^Q 










T 



M 

FIG. 24. 



Then the area PLNR lies between that of the trapezium SLNT 
and the sum of the areas of the trapezia PLMQ, QMNR. 



That 



Ca+k 

is, F(x)di 

J a 



ix lies between 

hF{a + \h)=A, say, 
and lh{F(a)+2F(a + lh)+F(a+h)}=B, say. 

In the figure F"(x) is positive and A is the lower limit, B the 
upper. If F"(x) were negative, A would be the upper limit and B 
the lower. 



* Phil. Mag., June 1919. Most of this paper is reproduced here. 



106 DIFFERENTIAL EQUATIONS 

As an approximation to the value of the integral it is best to 
take, not the arithmetic mean of A and B, but %B+^A, which is 
exact when PQR is an arc of a parabola with its axis parallel to the 
axis of x. It is also exact for the more general case when 

F (x) = a + bx + ex 2 + ex 3 , 
as is proved in most treatises on the Calculus in their discussion of 
Simpson's Rule. 

92. Extension of preceding results to functions denned by differential 
equations. Consider the function denned by 

^ =»/(*» y)> y= b wnen * = a ; 

where f(x, y) is subject to the following limitations in the range of 
values a to a + h for x and 6 - h to b + h for y. It will be seen from 
what follows below that the increment of y is numerically less than h, 
so that all values of y will fall in the above range. The limitations 
are : 

(1) f(x, y) is finite and continuous, as are also its first and second 
partial differential coefficients. 

(2) It never numerically exceeds unity. If this condition is not 
satisfied, we can generally get a new equation in which it is satisfied 
by taking y instead of x as the independent variable. 

(3) Neither cPy/dx 3 nor df/dy changes sign. 

" Let m and M be any two numbers, such that 

Then if the values of y when x is a +\h and a + h are denoted by 
b +j and b + h respectively,* 

-^h^lmh<j<^Mh^ih, (1) 

and -&= mh<k<Mh^h (2) 

We shall now apply the formulae of the last article, taking y to 
be the same function as that defined by 

Ca+x 

y = b + \ F(x)dx, 

J a 

Ca+h 

so that &=| F(x)dx. 

We have to express the formulae in terms of / instead of F. 
Now, F(a) =the value of dy/dx when x = a, 
so that F(a)=f(a, b). 

*The following inequalities hold only if h is positive. If h is negative, they 
must be modified, but the final result stated at the end of this article is still true. 



NUMERICAL APPROXIMATIONS 107 

Similarly, F(a + \h) =f(a + \h, b +j), 

and F(a + h)=f(a+h, b + k). 

Now, if df/dy is positive, so that/ increases with y, the inequalities 
<1) and (2) lead to 

f{a+\h, b+\mh)<f(a + \h, b+j)<f(a+\h, b+\Mh), (3) 

and f(a+h,b+mh)<f(a + h,b+k)<f(a + h,b+Mh); (4) 

while if df/dy is negative, 

/(a + \h, b + \mh) >f(a + %h, b +j)>f(a + \h, b + \Mh), ... (5) 

and f(a+h,b+mh)>f(a+h,b+k)>f(a + h,b+Mh) (6) 

Thus if F" (x) — dPy/dx 3 is positive and df/dy is also positive, the 
result of Art. 91, 

A<k<B, 

may be replaced by p<k<Q, (7) 

where p = hf(a + \h, b + \mh) 

and Q = \h{f{a, b) + 2f(a + ±h,b+ \Mh) +f{a + h,b+Mh)}; 

while if F" (x) is positive, and df/dy is negative, 

P<k<q, (8) 

where P = hf(a + \h, b + \Mh) 

and q = \h{f{a, b) + 2f(a + hh, b + ^mh) +f(a + h, b+ mh)}. 
Similarly, if F" (x) and df/dy are both negative, 

p>k>Q, (9) 

while if F" {x) is negative and df/dy positive, 

P>k>q (10) 

These results may be summed up by saying that in every case 

{subject to the limitations on/ stated at the beginning of this article) 

k lies between the greatest and least of the four numbers p, P, q, and Q. 

As an approximate formula we use k = %B+^A, replacing B by 

Q or q, and A by p or P. 

93. Application to a numerical example. Consider the example 
selected by Runge and Kutta to illustrate their methods, 

-t~=- ; w = lwhena;=0. 

ax y + x a 

It is required to find the increment k of y when x increases by 
0-2. Here f(x, y)=(y -x)/(y +x). This function satisfies the con- 
ditions laid down in the last article.* 

WetakeM = l, m = (l-0-2)/(l-2+0-2)=4/7. 

* As f (x, y) is positive, y lies between 1 and 1-2. When finding M and m we 
always take the smallest range for y that we can find 



108 DIFFERENTIAL EQUATIONS 

Then p =0-1654321, 

P =0-1666667, 
9=0-1674987, 

£=0-1690476. 
Thus h lies between p and Q. Errors. 

|Q + *p =0-1678424, 00000007 

Kutta's value 0-1678449, 0-0000032 

Kunge's value 0-1678487, 0-0000070 

Heun's value 0-1680250, 0-0001833 

The second, third, and fourth of these were calculated by Kutta. 
Now this particular example admits of integration in finite terms, 
giving 

log {x 2 +y 2 )-2 tan- 1 (x/y) =0. 

Hence we may find the accurate value of k. 
Accurate value =0-1678417. 
Thus in this example our result is the nearest to the accurate 
value, the errors being as stated above. 

We may also test the method by taking a larger interval h = \. 
Of course a more accurate way of obtaining the result would be to 
take several steps, say ^=0-2, 0-3, and finally 0-5, as Runge does. 

Still, it is interesting to see how far wrong the results come for 
the larger interval. 

We take M = 1, m = (l -l)/(2 +1) =0. 

Then iQ+iP =0-50000. 

True value =0-49828, Errors. 

Kutta's value =0-49914, 0-00086 

Our value =0-50000, 0-00172 

Heun's value =0-51613, 0-01785 

Runge's value =0-52381, 0-02553 

This time Kutta's value is the nearest, and ours is second. 



CHAPTER IX 

SOLUTION IN SERIES. METHOD OF FROBENIUS 

94. In Chapter VII. we obtained the solution of several equations 
of the form d 2 y n dy n A 

where P and Q were functions of x. 

In every case the solution was of the form 
y=af(x)+bF(x), 
where a and b were arbitrary constants. 

The functions f(x) and F(x) were generally made up of integral 
or fractional powers of x, sines and cosines, exponentials, and 
logarithms, such as 

(1 +2x)e x , sin x + x cos x, x*+x , x+\ogx, e' : . 

The first and second of these functions can be expanded by 
Maclaurin's theorem in ascending integral powers of x ; the others 
cannot, though the last can be expanded in terms of 1/x. 

In the present chapter, following F. G. Frobenius,* of Berlin, we 
shall assume as a trial solution 

y = x° (a + a x x + ag? + ... to inf.), 
where the a's are constants. f 

The index c will be determined by a quadratic equation called 
the Indicial Equation. The roots of this equation may be equal, 
different and differing by an integer, or different and differing by a 
quantity not an integer. These cases will have to be discussed 
separately. 

The special merit of the form of trial solution used by Frobenius 
is that it leads at once to another form of solution, involving log x, 
when the differential equation has this second form of solution. 

* Crelle, Vol. LXXVL, 1873, pp. 214-224. 

t In this chapter suffixes will not be used to denote differentiation. 

109 



110 DIFFERENTIAL EQUATIONS 

1 

As such a function as e x cannot be expanded in ascending powers 
of x, we must expect the method to fail for differential equations 
having solutions of this nature. A method will be pointed out by 
which can be determined at once which equations have solutions of 
Frobenius' forms (regular integrals) and for what range of values 
of x these solutions will be convergent. 

The object of the present chapter is to indicate how to deal 
with examples. The formal proofs of the theorems suggested will 
be given in the next chapter. 

Among the examples will be found the important equations of 
Bess*l,* Legendre, and Riccati. A sketch is also given of the Hyper- 
geometric or Gaussian equation and its twenty-four solutions. 

95. Case I. Roots of Indicial Equation unequal and differing by a 
quantity not an integer. Consider the equation 

v* + ^&- d £- 6 *y=° w 

Put z = X s (a + a x x + ag? + ...), where a =l=0, giving f 

dz 

j- =a cx c - x + a x (c + l)x° + a 2 (c + 2)x° +1 + ... , 

d 2 z 
and j^ = a c(c -\)x°-* -i-a^c + l)cx c ~ 1 +a 2 (c +2)(c + l)z? + ... . 

Substitute in (1), and equate the coefficients of the successive 
powers of x to zero. 

The lowest power of x is a? -1 . Its coefficient equated to zero gives 

a {2c(c-l)-c}=0, 

i.e. c(2c-3)=0, : (2) 

as a =f= 0. 

* Friedrich Wilhelm Bessel, of Minden (1784-1846), was director of the obser- 
vatory at Konigsberg. He is best known by " Bessel's Functions." 

Adrian Marie Legendre, of Toulouse (1752-1833), is best known by his "Zonal 
Harmonics" or "Legendre's Coefficients." He also did a great deal of work on 
Elliptic Integrals and the Theory of Numbers. 

Jacopo Francesco, Count Riccati, of Venice (1676-1754), wrote on " Riccati's 
Equation," and also on the possibility of lowering the order of a given differential 
equation. 

Karl Friedrich Gauss, of Brunswick (1777-1856), "the Archimedes of the 
nineteenth century," published researches on an extraordinarily wide range of 
eubjects, including Theory of Numbers, Determinants, Infinite Series, Theory of 
Errors, Astronomy, Geodesy, and Electricity and Magnetism. 

t It is legitimate to differentiate a series of ascending powers of x term by term 
in this manner, within the region of convergence. See Bromwich, Infinite Series, 
Art. 52. 



SOLUTION IN SERIES 111 

(2) is called the Indicial Equation. 

The coefficient of af equated to zero gives 

o,{2(c + l)c-(c + l)}=0, i.e. Oj-0 .....(3) 

The coefficient of x? +1 has more terms in it, giving 

a 2 {2(c + 2)(c + l)-(c+2)}+a {c(c-l)-6}=0, 
i.e. a 2 (c+2)(2c + l)+a (c + 2)(c-3)=0, 

i.e. a 2 (2c + l)+a (c-3)=0 (4) 

Similarly, a 3 (2c+3)+a 1 (c-2)=0, (5) 

a 4 (2c + 5)+a 2 (c-l)=0, .'. (6) 

and so on. 

From (3), (5), etc., 0=a 1 =a 3 =a 5 = ... =a 2n+v 
From (4), (6), etc., 

a 2 _ c - 3 a 4 _ c - 1 

a 2c + 1' a 2 _ 2c + 5' 

Og _ c + 1 o 2n c + 2w - 5 

a 4 ~ 2c +9' a 2n _ 2 2c + 4ri-3* 

But from (2), c=0orf. 

Thus, if c=0, 

z=a\l +3x 2 +=£* -y~x? +Wr2?--- \ =au, say, 
replacing a by a ; and if c=f, 

z=bx { 1 +3 a;2 -87i6 a;4 + 8.16.24 a;6 ~8.16.24.32 :r8 -J 
= bv say, replacing a (which is arbitrary) by b this time. 
Thus y=au + bv is a solution which contains two arbitrary con- 
stants, and so may be considered the complete primitive. 

In general, if the Indicial Equation has two unequal roots a and (3 
differing by a quantity not an integer, we get two independent solutions 
by substituting these values of c in the series for z. 

Examples for solution. 

(1 > <-.2 + $ + »- a (2) «-tt-*»3+a-)g+«i-a 

(4) Bessel's equation of order n, taking 2n as non-integral, 

x 2 ^| + x^ + (x 2 -« 2 )y = 0. 
ax 2 ax 



112 DIFFERENTIAL EQUATIONS 

96. Convergence of the series obtained in the last article. It is 

proved in nearly every treatise on Higher Algebra or Analysis that 
the infinite series %+w 2 + w 3 + ... is convergent if 



Lt 



l n+l 



<1. 



Now in the series we obtained u n — a 2 n-2?f +2n ~ 2 > *'•& 



"'n u '2n-2 

_ c + 2n-5 2 • 
~~~2c + 4n-3 X ' 

and the limit when w->oo is - \x z , independent of the value of c. 

Hence both series obtained are convergent for | x | < \/2. 

It is interesting to notice that if the differential equation is 
reduced to the form 

giving in our example p(x) = K — -s, 

o^c 

and q(x)=^—— , 

n ' 2 + x 2 

■p(x) and q(x) are expansible in power series which are convergent 
for values of x whose modulus | x | < \/2. 

That is, the region of convergence is identical in this example 
with the region for which p (x) and q (x) are expansible in convergent 
power series. We shall show in Chap. X. that this theorem is true 
in general. 

Examples for solution. 

Find the region of convergence for the solutions of the last set of 
examples. Verify in each case that the region of convergence is identical 
with the region for which p(x) and q(x) are expansible in convergent 
power series. 

97. Case II. Roots of Indicial Equation equal. Consider the 
equation 

<.-*-)g + <l-lte)g-*-0. 

Put z=x c (a + a 1 x+a 2 x 2 + ...), 

and after substituting in the differential equation, equate coefficients 
of successive powers of x to zero just as in Art. 95. 






SOLUTION IN SERIES 

We get a Q {c(c-l)+c}=0, 

i.e. c 2 =0, 

a 1 {(c + l)c + c + l}-a {c(c-l)+5c + 4}=0, 

i.e. flr 1 (c + l) a -o (c+2) a =0, 

a 2 (c + 2) 2 -a 1 (c+3) 2 =0, 

a 3 (c + 3) 2 -a 2 (c + 4) 2 =0, 



113 



and so on. 
Hence 



•(1) 

•(2) 

(3) 

•(4) 



•-^K^'-'C-SD 






is a solution if c=0. 

This gives only one series instead of two. 

But if we substitute the series in the left-hand side of the dif- 
ferential equation (without putting c=0), we get the single term 
ffocV. As this involves the square of c, its partial differential 
coefficient with respect to c, i.e. 2a cx° + a^x* log x will also vanish 
when c=0. 

That is, 

d 



dc 



d 2 d 

( X ~ X ^ dx 2 + (1 " 5a;) dx ~ 4 J ^ = 2a ° cxC + ttocV lo § x - 



As. the differential operators are commutative, this may be 
written 

r d 2 d "1 dz 

1 {X ~ x2) dx 2 + {1 ~ 5x) d^- i ]Fc = 2a ° CxC + «<?* lo 8 x - 

Hence ^ is a second solution of the differential equation, if c is 

put equal to zero after differentiation. 
Differentiating, 

l Z = z\o %X + a^[2( C -^). r - 1 X+ 2( C+S ).~ 2 x * 

■c+i\ -3 „ \ 



+ 2 



c + i/ (c + iy 

Putting c=0 and a Q =a and b respectively in the two series, 
z - a{l 2 + 2 2 x + 3 2 x 2 + W + 5 2 x 4 + ...}= au, say, 
dz 

dc 



and .-=6wloga;-26{l .2x+2 .3x 2 +3 . Ax* + ...}=bv, say. 



The complete primitive is au + bv. 

P.D.E. H 



114 DIFFERENTIAL EQUATIONS 

In general, if the Indicial Equation has two equal roots c = a, 

we get two independent solutions by substituting this value of c in z and 

dz 

~-, The second solution will always consist of the product of the 

first solution (or a numerical multiple of it) and log a;, added to 
another series. 

Reverting to our particular example, consideration of p (x) 
and q (x), as in Art. 96, suggests that the series will be convergent 
for | x | < 1. It may be easily shown that this is correct. 

Examples for solution. 

(i) c-^g + (i-*»2-,-a 

(2) Bessel's equation of order zero 

(4) 4(^-^)g + 8^|-y=0. 

98. Case III. Roots of Indicial Equation differing by an integer, 
making a coefficient of z infinite. Consider Bessel's equation of order 

unit F' 9 &y fy 1 2 n a 

dx 2 dx v /y 

If we proceed as in Art. 95, we find 

a {c(c-l)+c-l}=0, 

i.e. c 2 -l=0, (1) 

^{(c + l^-lHO, 

i.e. Oi^O, (2) 

a 2 {(c+2) 2 -l}+a =0, (3) 

and a n {{c+n) 2 -l}+a n _ 2 =0, (4) 

giving 



= a £ c -U -—. — ; w ~<^x 2 +- 



($ + l)(^+3) (c + l)(c+3) 2 (c+5) 



a^ + .-.i- 



(c + l)(c+3) 2 (c+5) 2 (c+7) 
The roots of the indicial equation (1) are c = 1 or - 1. 
But if we put c = - 1 in this series for z, the coefficients become 
infinite, owing to the facVor (c + 1) in the denominator. 



SOLUTION IN SERIES 115 

To obviate this difficulty replace * a by (c + l)k, giving 

V (c+3) (c+3) 2 (c+5) 

" (c+3) 2 (c + 5) 2 (c + 7) a;6 ' + " "/' (5) 

and a; 2 ^ 2 +^ + (cc 2 -l)2 = ^(c + l)(c 2 -l)=^(c + l) 2 (c-l). 

Just as in Case II. the occurrence of the squared factor (c + 1) 2 

dz 
shows that =-, as well as z, satisfies the differential equation when 

c = - 1. Also putting c = 1 in z gives a solution. So apparently we 
have found three solutions to this differential equation of only the 
Second order. 

On working them out, we get respectively 

fari{-^B»+^s*- 2a l 2 6 x« + ...j=Jcu, say, 
ku log x +&*r 1 jl +- a;2 ___ (_ + _) ^ 

+ 2 2 rlo(i + i + o) :c6+ -) = ^' say ' 

and fee { 2 -4^ + 4276^ - 42 . 6 a , g ^ 6 + •••} = A ™> sa 7- 

It is obvious that w= - 4w, so we have only found two linearly 
independent solutions after all, and the complete primitive is au + bv. 
The series are easily proved to be convergent for all values of x. 

The identity (except for a constant multiple) of the series obtained 
by substituting c ■= - 1 and c = 1 respectively in the expression for z 
is not an accident. It could have been seen at once from relation (4), 
a„{(c+?i) 2 -l}+a„_ 2 =0. 

If c = l, this gives a n {(l +n) 2 -1} + a„_ 2 =0 (6) 

Ifc=-1, a n {(-l+w) 2 -l}+a n _ 2 =0; 

hence replacing n by n + 2, 

a n+2 {(l+n)*-l}+a n =0 (7) 

[ a n+2 _ a n /o\ 

a n Jc=-l L#„_ 2 J c =i 

As [£](;=_;! has x~ x as a factor outside the bracket, while [z] c=1 has 
x, relation (8) really means that the coefficients of corresponding 

Of course the condition a ^0 is thus violated ; we assume in its place that 
khO. 



Thus 



116 DIFFERENTIAL EQUATIONS 

powers of x in the two series are in a constant ratio. The first series 
apparently has an extra term, namely that involving x~ x , but this 
conveniently vanishes owing to the factor (c + 1). 

In general, if the Indicia! Equation has two roots a and /3 (say 
a > /3) differing by an integer, and if some of the coefficients of z become 
infinite when c=/3, we modify the form of zby replacing a by k(c-/3). 
We then get two independent solutions by putting c=fi in the modified 

dz 
form of z and ~-. The result of putting c=a in z merely gives a 

numerical multiple of that obtained by putting c =j3. 

Examples for solution. 

(1) Bessel's equation of order 2, 

(3)*(l-»)g-(l+3*)g-!f-a. 

(4) (x + x 2 + a?)^ + 3x 2 ^--2y = 0. 
ax 1 dx 

99. Case IV. Roots of Indicial Equation differing by an integer, 
making a coefficient of z indeterminate. Consider the equation 



d *y ,oJy 

dx'' 
Proceeding as usual, we get 



.-****.- I 



c(c-l)=0, (1) 

a x (c + l)c=0, (2) 

a 2 (c+2)(c + l)+a {-c(c-l)+2c + l}=0, (3) 

« 3 (c+3)(c+2)+a 1 {-(c + l)c+2(c + l)+l}=0, (4) 

and so on. 

(1) Gives c=0 or 1. 

The coefficient of a x in (2) vanishes when c =0, but as there is no 
other term in the equation this makes a x indeterminate instead of 
infinite. 

If c-1, «i =0. 

Thus, if c=0, from equations (3), (4), etc. 

2a 2 + « =0, 
6a 3 +3a 1 =0, 
12a 4 +3a 2 =0, 
etc., 



SOLUTION IN SERIES 117 

giving [z] c =o = «o(l-2 a;2+ 8 a;4+ 80 



a x {x-\ 



. 1 , 3 7 1 



This contains two arbitrary constants, so it may be taken as the 
complete primitive. The series may be proved convergent for 
\x\<l. 

But we have the other solution given byc = l. Working out 
the coefficients, 



[2] c=1 = «oa;| 



1 "2 x2 + 40 a;4+ 560 a;6 - 



that is, a constant multiple of the second series in the first solution. 

This could have been foreseen from reasoning similar to that in 
Case III. 

In general, if the Indicial Equation has two roots a and /3 (say 
a >/3) differing by an integer, and if one of the coefficients of zjwcomes 
indeterminate when c = /3, the complete primitive is given by putting 
c=/3 in z, which then contains two arbitrary constants. The result oj 
putting c = a in z merely gives a numerical multiple of one of the series 
contained in the first solution. 

Examples for solution. 

(1) Legendre's equation of order unity, 

(2) Legendre's equation of order n, 

(l-x«)^-2*| + n(n + l)t, = 0. 

(3) + ^=0. (4) (2 + x*) d ^ 2 + xf x + (l + x)y = 0. 

i 
100. Some cases where the method fails. As e x cannot be expanded 

in ascending powers of x, we must expect the method to fail in 

some way when the differential equation has such a solution. To 

construct ah example, take the equation ~ -y=0, of which e z 
and e~ z are solutions, and transform it by putting z =-. 
We have dy_dx dy_ 1 dy _ dy 

>»e nave 7 - — 7 , — „ 7 — jj i 

dz dz dx z l dx dx 

and ^/ = <te *(<k) = - X 2<L( - x * d ' f )=x* d * y + 2%» (Jy . 
dz 2 dz dx\dz/ dx\ dx) dx 2 ^ dx 



118 DIFFERENTIAL EQUATIONS 

Hence the new equation is 

If we try to apply the usual method, we get for the indicial 

equation, -a =0, which has no roots,* as by hypothesis a =/=0. 

Such a differential equation is said to have no regular integrals 

i _i 

in ascending powers of x. Of course e x and e x can be expanded in 

powers of - . 

The examples given below illustrate other possibilities, such as 
the indicial equation having one root, which may or may not give 
a convergent series. 

It will be noticed that, writing the equation in the form 

in every case where the method has succeeded p(x) and q(x) have 
been finite for x=0, while in all cases of failure this condition is 
violated. 

For instance, in the above example, 
p(x)=-2, 

q(x)= — - 2 , which is infinite if x=0. 

Examples for solution. 

(1) Transform Bessel's equation by the substitution x = l/z. 
Hence show that it has no integrals that are regular in descending 

powers of x. 

(2) Show that the following equation has only one integral that is 
regular in ascending powers of x t and determine it : 

(3) By putting y = vx 2 (l +2x) determine the complete primitive of 
the previous example. 

(4) Show that the following equation has no integral that is regular 
in ascending powers of x, as the one series obtainable diverges for all 
values of a;: My dy 

(5) Obtain two integrals of the last example regular in descending 
powers of x. 

* Or we may say that it has two infinite roots. 



SOLUTION IN SERIES 119 

(6) Show that the following equation has no integrals that are 
regular in either ascending or descending powers of x : 

s«(l - x 2 ) Pi + 2x* P - (1 - x 2 fy =0. 
ax 2 ax 

[This is the equation whose primitive is ae x+x ~* + be~ x ~ x '^.] 



MISCELLANEOUS EXAMPLES ON CHAPTER IX. 

(1) Obtain three independent solutions of 

(2) Obtain three independent solutions, of the form 

dz . d 2 z 

z > dc> and a?' 

of the equation x 2 ~ + Sx-r^ + (1 - x) -?■ - y = 0. 
T ax 2 ax 2 ax 

(3) Show that the transformation y — j-j- reduces Riccati's equation 

^ + by 2 = cx m 

(L v 
to the linear form -=-= - bcvx m =0. 

dx- 

(4) Show that if y is neither zero nor an integer, the Hyper geometric 
Equation 1% j 

x(l-x)^ 2 + {y-(a + {3 + l)x}£-apy = 

has the solutions (convergent if \x\ < 1) 

F(a,/3,y,x) and &-*F(a-y + \ t /3-y + l, 2-y, x), 

where F(a, f3, y, x) denotes the Hypergeometric Series 

. , a/3 r , a(a + l)i8(i8 + l) ra , a(a + l)(a + 2) j 8(i8 + l)( i 9 + 2) 
l.y 1.2.y(y + l) 1 . 2 . 3 . y(y + l)(y + 2) 

(5) Show that the substitutions a? = 1 - z and x = 1/z transform the 
hypergeometric equation into 

z(l- 2 )g+{a + /3 + l-y-(a + /3 + l)2}!-«/fy = 

and z 2 (l-z)0 + z{(l-a-/3)-(2-y)4^ + a/fy = O 

respectively, of which the first is also of hypergeometric form. 



120 DIFFERENTIAL EQUATIONS 



■ 



Hence, from the last example, deduce that the original equation has 
the additional four solutions : 

F(a,0, a + p + l-y, 1-x), 
(\-x)y—PF(y-l3,y-a.l+y-a-^,l-x), 
x~ a F(a, a + l-y, a + l-/8, or 1 ), 
and x-PF(P, p + l-y,p + l- a , ar 1 ). 

(6) Show that the substitution y — (1 -x) n Y transforms the hyper- 
geometric equation into another hypergeometric equation if 

n=>y-a-j3. 
Hence show that the original equation has the additional two 
solutions : (i _ x )y-«-^( y _ aj y _^ y> x ) 

and x 1 -y(l-x)y*-^F(l-a, l-/3,2-y,x). 

[Note. — Ex. 5 showed how from the original two solutions of the 
hypergeometric equation two others could be deduced by each of the 
transformations x = l-z and x = l/z. Similarly each of the three 

transformations x = = , x = — -, x = . 'gives two more, thus making 

1 -Z 2-1 Z & ° 

twelve. By proceeding as in Ex. 6 the number can be doubled, giving 
a total of twenty-four. These five transformations, together with the 
identical transformation x = z, form a group ; that is, by performing two 
such transformations in succession we shall always get a transformation 
of the original set.] 

(7) Show that, unless 2w is an odd integer (positive or negative), 
Legendre's equation 

(l-**) d ^-2x d £ + n(n + l)y=0 

has the solutions, regular in descending powers of x, 
x- n ~ 1 F{\n + \, bi + 1, n+f, x~ 2 ), 
x n F(-%n, \-ln, \-n, x~ 2 ). 
[The solution for the case 2n= - 1 can be got by changing x into 
x- 1 in the result of Ex. 4 of the set following Art. 97.] 

(8) Show that the form of the solution of Bessel's equation of 
order n depends upon whether n is zero, integral, or non-integral, 
although the difference of the roots of the indicial equation is not n 
but 2n. 



* CHAPTER X 

EXISTENCE THEOREMS CF PICARD, CAUCHY,f AND 
FROBENTUS 

101. Nature of the problem. In the preceding chapters we have 
studied a great many devices for obtaining solutions of differentia] 
equations of certain special forms. At one time mathematicians 
hoped that they would discover a method for expressing the solution 
of any differential equation in terms of a finite number of known 
functions or their integrals. When it was realised that this was 
impossible, the question arose as to whether a differential equation 
in general had a solution at all, and, if it had, of what kind. 

There are two distinct methods of discussing this question. 
One, due to Picard, has already been illustrated by examples 
(Arts. 83 and 84). We obtained successive approximations, 
which apparently tended to a limit. We shall now prove that 
these approximations really do tend to a limit and that 
this limit gives the solution. Thus we shall prove the exist- 
ence of a solution of a differential equation of a fairly general 
type. A theorem of this kind is called an Existence Theorem. 
Picard's method is not difficult, so we will proceed to it at once 
before saying anything about the second method. It must be 
borne in mind that the object of the present chapter is not to 
obtain practically useful solutions of particular equations. Our 
aim now is to prove that the assumptions made in obtaining 
these solutions were correct, and to state exactly the conditions 
that are sufficient to ensure correctness in equations similar to 
those treated before, but generalised as far as possible. 

* This chapter should be omitted on a first reading. 

t Augustin Louis Cauchy. of Paris (1789-1857), may be looked upr n as the 
creator of the Theory of Functions and of the modern Theory of Differential Equa- 
tions. He devised the method of determining definite integrals by Contour 
Integration. 

121 



122 DIFFERENTIAL EQUATIONS 

102. Picard's method of successive approximation. If -^ =f(x, y) 

and y = b when x = a, the successive approximations for the value 
of y as a function of x are 



o + f f(x, b)dx=y v ssij, 

b + I f(x, y 2 )dx =y 3 , say, and so on. 



We have already (Arts. 83 and 84) explained the application of 
this method to examples. We took the case where f(x, y)=x+y 2 , 
b=a=0, and found 

These functions appear to be tending to a limit, at any rate f&r 
sufficiently small values of x. It is the purpose of the present 
article to prove that this is the case, not merely in this particular 
example, but whenever f(x, y) obeys certain conditions to be 
specified. 

These conditions are that, after suitable choice of the positive 
numbers h and k, we can assert that, for all values of x between 
a-h and a + h, and for all values of y between b-Jc and b+k, we 
oan find positive numbers M and A so that 

(i) \f(x,y)\<M, 

(ii) \f(x,y)-f(x,y')\<A\y-y'\, y and y' being any two 
values of y in the range considered. 

In our example /(cc, y) =x+y 2 , condition (i) is obviously satisfied, 
taking for M any positive number greater than (h + k 2 ). 

Also \(x + y 2 )-(x + y' 2 )\=\y + y'\\y-y'\<2k\y-y'\, 
so condition (ii) is also satisfied, taking A =2k. 

Returning to the general case, we consider the differences between 
the successive approximations. 

V\ ~ ° = f( x > b)dx, by definition, 

J a 

but \f(x, b)\<M, by condition (i), 



■so 



I V\ ~ & I < I Mdx 

I J a 



i.e. <M\x-a\<Mh (1) 



EXISTENCE THEOREMS 123 



Also y2~yi=b + \ f(x, yjdx-b - I f(x, b) dx, by definition, 

J a Ja 

= \ {f{x i y 1 )-f(x i b)}dx; 

J a 

but | f(x, y x ) -f(x, b) | < A | y x - b | , by condition (ii), 

<AM\x-a\, from (1), 

$o \y 2 -y 1 \< II AM(x-a)da 

I Ja 



i.e. <|M(j-a)k|M 2 ,...(2) 



Similarly, | y w -*,„_, | <-, MA»-*K n (3) 



1 

ni 
Now the infinite series 

1 «._ „ M 



is convergent for all values of h, A, and M. 
Therefore the infinite series 

b + {y 1 -b)+(y 2 -y 1 ) + ...+ (y n - y n _ x ) + ..., 

each term of which is equal or less in absolute value than the corre- 
sponding term of the preceding, is still more convergent. 
That is to say that the sequence 

yi=b + {yx-b), 

y2 = b + (y 1 -b) + (y 2 -y l ), 

and so on, tends to a definite limit, say Y(x), which is what we 
wanted to prove. 

We must now prove that Y satisfies the differential equation. 

At first sight this seems obvious, but it is not so really, for we 
must not assume without proof that 

Lt f(x,y n _ 1 )dx=\ f{x, Lt y n ^)dx. 

The student who understands the idea of uniform convergence 
will notice that the inequalities (1), (2), (3) that we have used to 
prove the convergence of our series really prove its uniform con- 
vergence also. If, then, f(x, y) is continuous, y v y 2 , etc., are 
continuous also, and Y is a uniformly convergent series of con- 
tinuous functions ; that is, Y is itself continuous,* and Y - </„_! 
tends uniformly to zero as n increases. 

Hence, from condition (ii), f(x, Y)-f(x, y n _^) tends uniformly 
to zero. 

* See Bromwich's Infinite Series, Art. 45. 



J a 



124 DIFFERENTIAL EQUATIONS 

From this we deduce that 

Y) -f(x, i/n-x)} tends to zero. 
Thus the limit of the relation 

y n = b + \ f(x, y n . x )dx 

J a 

is Y=b + \ X f{x,Y)dx; 

J a 

dY 
therefore* -=—=f(x, Y), and Y =6 when x = a. 

This completes the proof. 

103. Cauchy's method. Theorems on infinite series required. 

Cauchy's method is to obtain an infinite series from the differential 
equation, and then prove it convergent by comparing it with another 
infinite series. The second infinite series is not a solution of the 
equation, but the relation between its coefficients is simpler than 
that between those of the original series. Our first example of this 
method will be for the simple case of the linear equation of the first 
order dy , . 

f x =vw-y- 

Of course this equation can be solved at once by separation of 
the variables, giving 

logy=c + I p(x)dx. 

However, we give the discussion by infinite series because it is 
almost exactly similar to the slightly more difficult discussion of 

dx^ {x) -dx +q{x) - y > 
and other equations of higher order. 

We shall need the following theorems relating to power series. 
The variable x is supposed to be complex. For brevity we shall 
denote absolute values by capital letters, e.g. A n for \a n \. 

CO 

(A) A power series "V ««#" is absolutely convergent at all 

o 
points within its circle of convergence \x \=R. 

(B) The radius R of this circle is given by 
provided that this limit exists. 

* When differentiating the integral, the student should remember that the 
integral varies solely in consequence of the variation of its upper limit. 



EXISTENCE THEOREMS 125 

^) Tx (i>«* n ) - S ^n*"- 1 , Within |*|-$. 

(D) If we have two power series, then for points within the 
circle that is common to their circles of convergence, 

(00 \ / 00 \ 00 

2 a * xn ) (S ^"J = 2 ( a « 6 o +««-A + ... + %K)% n - 
o ' v o 7 u 

oo oo 

(£") If ^ a n £ n = V b n x n for all values of x within the circle 
o o 

\x\ = R, then a n = b n . 

(F) A n < MR~ n , where M exceeds the absolute value of the 
sum of the series at points on a circle |a;|=J2 on which the series 
is convergent. 

Proofs of these theorems will be found in Bromwich's Infinite 
Series : 

A in Art. 82, 

B is an obvious deduction from D'Alembert's ratio test, Art. 12, 
C in Art. 52, 
D » 54, 
E „ 52, 
F „ 82. 

Two theorems on uniform convergence will be required later on, 
but we will defer these until they are needed. 

dy 
104. * Convergence of the solution in series of — =yp (x). Let 

n 

00 

p(x) be capable of expansion in a power series y]p n x n which is 

o 
convergent everywhere within and on the circle | » | = R. We shall 

00 

prove that a solution y = S] a n x n can be obtained which is 

o 
convergent within this circle. 

Substituting in the differential equation, we obtain 

00 GC GO 

V nanX"- 1 - 2 a n x n 2 p n x n (Theorem C) 



00 

* 2 ( a nPo + «n-i?i + a n -zP2 + • • • + «o?n) x n . (Theorem D) 



Equating the coefficients of sc n_1 , (Theorem E) 

na n =a n _ 1 p + a n _ 2 p 1 +a n _ 3 p 2 +... + a p n _ l (1) 

* Revise Art. 7 before reading the following. 



126 DIFFERENTIAL EQUATIONS 

Hence for the absolute values of the a's and/'s, denoted by the 
corresponding capital letters, we get 

nA n zc A n _ 1 P Q + A n _ t P 1 + A n _ z P 2 + ... + A Q P n ^ (2) 

Let M be a positive number exceeding the absolute value of 
p (x) on the circle \x\=R, 

then P n <MR~ n ; ..... (3) (Theorem F) 

therefore, from (1) and (3), 

M 

A n < — (A n _ 1 + A n _ 2 R- 1 +A n _ 3 R-* + ...+A R- n +i) (4) 

to 

Define B n (n > 0) as the right-hand side of (4), and define 
B as any positive number greater than A ; then A n < B n . 

M 

But -(A n _ 1 + A n _ 2 R-i+A n _sR-* + ... +A R-"+i) 

M n — 1 M 

=— A n _ x + nR nl (A n _ 2 + A n _ 3 R-i + ... +A R~ n +*). 

Hence, defining B n as above, 

_M (n-l)B n _ r 

ix ' < \n + RJ Bn ~ 1 ' as An - X < Bn - X ' 

therefore _^L<^ + 1, 

B n _ 1 ^n + R' 

T , B n 1 
i.e. Lt r^-^ p- 

n— >-oo -Djj— 1 " 

Therefore the series ^ B n x n is convergent within the circle 
\x\=R. o (Theorem B.) 

00 

Still more therefore is the series y] a n x n convergent within the 

o 
same circle, since A n < B n . 

The coefficients a x , a 2 , ... can all be found from (1) in terms of 
the £>'s, which are supposed known, and the arbitrary constant a . 

105. Remarks on this proof. The student will probably have 
found the last article very difficult to follow. It is important not 
to get confused by the details of the work. The main point is this. 

We should like to prove that Lt . n ^R. Unfortunately the 

n->oo -fin— i 

relation defining the A's is rather complicated. We first simplify 
it by getting rid of the n quantities P , P lf ...P n .. 1 . Still the 



EXISTENCE THEOREMS 127 

relation is too complicated, as it involves n A' a. We need a simple 
relation involving only two. By taking a suitable definition 
of B n we get such a relation between B n and B n _ x , leading to- 

Lt ^-<R- 

We repeat that the object of giving such a complicated dis- 
cussion of a very simple equation is to provide a model which the 
student can imitate in other cases. 

Examples for solution. 

(1) Prove that, if p(x) and q(x) can be expanded in power series 
convergent at all points within and on the circle X = R, then a power 
series convergent within the same circle can be found in terms of the 
first two coefficients (the arbitrary constants) to satisfy 

g=^). ! + ^)*, 

[Here n (n-l)a n = (n- l)a n _ lPo + (n - 2)a n _ 2 p 1 + ...+ a x p n _ t 

+ a n-Z% + «n-S?l + • • • + «0?n-2- 

Hence, if M is any number exceeding the absolute values of both 
p(x) and q(x) at all points on the circle X = R, 

M 

A n < -{(A n _ } +A n _ 2 R-i + ... +A 1 R~"+*) 

+ (A n _ 2 + A n _ z R~i + . . . + A R-»+*)} 
<M(l+R)(A n _ 1 + A n _ 2 R-i + ...+A R-«+i). 

Define the right-hand side of this inequality as B n and then proceed 
as before.] 

(2) Prove similar results for the equation 

106. Frobenius' method. Preliminary discussion. When the 
student has mastered the last article, he will be ready for 
the more difficult problem of investigating the convergence of 
the series given by the method of Frobenius. In the preceding 
chapter (which should be thoroughly known before proceeding 
further), we saw that in some cases we obtained two series 
involving only powers of x, while in others logarithms were 
present. 

The procedure in the first case is very similar to that of the last 
article. But in the second case a new difficulty arises. The series 
with logarithms were obtained by differentiating series with 



128 DIFFERENTIAL EQUATIONS 

Tespect to a parameter c. Now differentiation is a process of taking 
a limit and the summation of an infinite series is another process 
of taking a limit. It is by no means obvious that the result will 
be the same whichever of these two processes is performed first, 
■even if the series of differential coefficients be convergent. 

However, we shall prove that in our case the differentiation is 
legitimate, but this proof that our series satisfy conditions sufficient 
to justify term-by-term differentiation is rather long and bewildering. 

To appreciate the following work the student should at first 
ignore all the details of the algebra, concentrating his attention on 
the general trend of the argument. When this has been grasped, 
he can go back and verify the less important steps taken for granted 
on a first reading. 

107. Obtaining the coefficients in Frobenius' series when the roots 
of the indicial equation do not differ by an integer or zero. Consider 
the expression 

. d 2 y . . dy . / dy d 2 y\ 

where p (%) and q (x) are both expansible in power series ^ p n x n 

CO o 

and V q n x n which are convergent within'and on the circle | x \ = R. 
o 
We are trying to obtain a solution of the differential equation 

♦(**£3)-« -w 

If y is replaced by x° j^ a n x n (with a =f 0), <p (x, y, J, ^Jj 
becomes ° 

^a n xP +n {(c + n)(c+n-l)-(c + n)p(x)-g(x)} 



oo 

= 2 9n^ +n , say, 



where g = a o M " 1 ) -PoC-<lo} 

and g n = a n {(c + n) (c +n - 1) -p {e+n)-q } 

-«n-i (Pi (c + n - 1) +fc} -a n _ 2 {p 2 (c+n-2) +q 2 } 
... -a (p n c+q n ). 
For brevity, denote 

c (c - 1) - p c-q Q by f(c), 
so that (c + n) (c + n - 1) - p (c + n) - q =f (c + n). 



EXISTENCE THEOREMS 129 

Then0 n =Oif 
a n Ac + n)=a n _ 1 {p 1 {c + n-l)+q 1 }+a n __ 2 {p 2 (c + n-2)+q 2 } 

+ ...+a (p n c+q n ) (2) 

If we can choose the a's so that all the #'s vanish, and if the 

00 

series ^ a n x n so obtained is convergent, a solution of (1) will have 

o 
been obtained. 

Now as a =^-0, g Q =0 gives 

c ( c -l)-^oC-? =0 (3) 

This is a quadratic equation in c, and is called the Indicial 
Equation. 

Let its roots be a and /3. 

If either of these values is substituted for c in the equations 
9i = ®> <72 =0 > 5 r 3 =0 > •••> values for a v a 2 , a 3 , ... are found in the form 

a n =aji n (c)/[f(c+n)f(c + n-l) .../(o + l)], (4) 

where h n (c) is a polynomial in c. The student should work out the 
values of a y and a 2 in full if he finds any difficulty at this point. 

The process by which a n is obtained from (2) involves division 
by / ( c + n )- This is legitimate only when / (c + n)=/=Q. 

Now as f(c) = (c-a) {c-/3), 

f(c + n) = (c +n - a) (c + n - /3), 

so f(a +n)=n(a+n-fi), (5) 

and f({3+n)=n(/3+n-a) (6) 

Thus, if a and /3 do not differ by an integer or zero, the divisors 
cannot vanish, so the above process for obtaining the «'s is satis- 
factory. 

108. Convergence of the series so obtained. Let M be a positive 
number exceeding the absolute values of p(x) and q(x) at all points 
on the circle | x \ =R. 

Then P»<MR-° 

and Q,<MR", 

so that | p s (c +n - s) + q„\ < M (C + n - s + l)R~ 3 . 

From these inequalities and from (2), 

A n < M{A n ^(C+n)R- 1 + ... +A Q {C + l)R- n }/F(c+n), ...(7) 
iay A n <B n , denoting the right-hand side of (7) by B„. This 
iefines B n if n>0. Define B as any positive number greater 
'"Jian A . This definition of B n gives 

B n+l F(c + n+ 1) -B n F(c +n)R~ 1 =A n M(C +n + l)R~ l 

<B n M(C + n + l)R~\ 



130 DIFFERENTIAL EQUATIONS 

so that *j*< r l e+ ?J M < 0+ * + 1 \ 
B n RF(c + n + l) 

i e < l(o+w)(c + n-l)-y (c+n) - g \+M( C+n + l) ^ 
R\{c+n + l)(c+n)-p (c+n + l)-q \ 

Now for large values of n the expression on the right approaches 
the value n2 i 

Rn 2 = R' 
Thus Lt-"^<C^. 

00 00 

Therefore the series V} 2?„a; n and still more the series V a n a: n 

U 

converges within the circle \x\ = R. 

Thus, when a and ft do not differ by an integer, we get two 
convergent infinite series satisfying the differential equation. 

109. Modification required when the roots of the indicial equation 
differ by zero or an integer. "When a and ft are equal, we get one 
series by this method. 

When a and ft differ by an integer, this method holds good 
for the larger one, but not for the smaller, f or if a - ft = r (a positive 
integer), then from (5) and (6) 

f(a +n)=n(a +n- ft) =n(n +r), 
but f(ft+n)=n(ft + n -a)=n(n-r), 

which vanishes when n = r, giving a zero factor in the denominator 
of a r when c=ft. As exemplified in Arts. 98 and 99 of the preceding 
chapter, this may give either an infinite or indeterminate value for 
some of the a's. This difficulty is removed by modifying the form 
assumed for y, replacing a by k(c-ft). This will make a , a lt ... , 
a,_, all zero and a n a r+1 , ... all finite when c is put equal to ft. This 
change in the form assumed for y will not alter the relation between 
the a's, and so will not affect the above investigation of convergence. 

110. Differentiation of an infinite series with respect to a parameter 
c, the roots of the indicial equation differing by an integer. In Art. 107 

we obtained an infinite series af"S]a n x n , where the a's are functions 

o 

of c. As in the preceding chapter, we have to consider the 

differentiation of this series with respect to c, c being put equal to 

the smaller root ft after the differentiation. 



EXISTENCE THEOREMS 131 

Now while this differentiation is being performed we may con- 
sider a; as a constant. The series can then be considered as a series 

of functions of the variable c, sayVi/^c), where 

= af+*aji n (c)/[f(c+n)f(c+n-l) .../(« + 1)1 from ( 4 )» 
where a Q = k(c-fi) and the factor (c-/3) is to be divided out if it 
occurs in the denominator. 

Now Qoursat (Cours d' Analyse, Vol. II. 2nd ed. p. 98) proves 
that if (i) all the xfs's are functions which are analytic and holo- 
morphic within a certain region bounded by a closed contour and 
continuous on this contour, and if (ii) the series of \//-'s is uniformly 
convergent on this contour, then the differentiation term by term 
gives a convergent series whose sum is the differential coefficient 
of the sum of the original series. 

For the definitions of holomorphic and analytic, see the beginning 
of Vol. II. of Goursat. It will be seen that the xfr's satisfy these 
definitions and are continuous as long as we keep away from values 
of c that make them infinite. These values are a -I, /3-1, a -2, 
j8 -2, etc. To avoid these take the region inside a circle of centre 
c = /3 and of any radius less than unity. 

We shall now prove that the series is uniformly convergent 
everywhere inside this region. This will prove it is uniformly 
convergent on the contour of a similar but slightly smaller region 
inside the first. 

Let s be a positive integer exceeding the largest value of C within 
the larger region. 

Then for all values of c within this region, for values of n exceed- 
ing 8, 
F (c + n) = \(c + n)(c +n -1) -p Q (c+n) -q \, by definition of F, 
^(C+n) 2 -{P + l){C+n)-Q , as \u-v\ > \ u \ - \ v\, 

> (n - s) 2 - (M + l)(s +n) - M, as P <M and Q <M, 

> n 2 +In+J, say, where / and J are independent of 

n, x, or c (8) 

For sufficiently great values of n, say n>m, the last expression 

is always positive. 

Let H denote the maximum value of 

M{A m _ x {C +m)R- 1 +A m _ 2 (C+m -l)R~ 2 + ... +A (C + 1 )R~ m ] (9) 

for all the values of c in the region. 



132 DIFFERENTIAL EQUATIONS 

Then if E m be any positive number greater than B m , and .-if, 
for values of n > m, E n be denned by 

F _ M{E n _ x {s +n)R~ 1 + ... E m (s + m + 1 ) fl-"+™} + HR~ n ^ 
n n*+In+J ' UU) 

an that F ME m (s+ m+l)R-i + HR-i 

so that E m+1 - {m + 1) 2 + I{m + 1)+J > 

which has a numerator greater than and a denominator less than 
those of B m+l , from (8), (9), and the definition of B n as the right- 
hand side of (7), we see that 

E m +1 > "m+V 

Similarly E n > B n for all values of n > m. 

J? 1 

From (10) we prove Lt -^ ±i =D- This piece of work is so 

similar to the corresponding work at the end of Art. 108 that we 
leave it as an exercise for the student. 

CO 

Hence ^ E n R x n is convergent if R X <R. 

in 

Therefore within the circle | x | = R\ and within the region 
specified for c, 

I a n x c + n | < A n R 1 t+n < B n R 1 s + n < E n R 1 ' +n . 

This shows that Ha n x c+n satisfies Weierstrass's M-test for uniform 
convergence (Bromwich, Art. 44), as R v s, and the £"s are all inde- 
pendent of c. 

This completes the proof that 2^ r j=2a„af +n satisfies all the 
conditions specified, so the differentiation with respect to c is now 
justified. This holds within the circle 1^1=^. We can take R x 
great enough to include any point within the circle |a;| =R. 

If the roots of the indicial equation are equal instead of differing 
by an integer, the only difference in the above work is that a is 
not to be replaced by k(c-(3), as no (c-/3) can now occur in the 
denominator of a n . 



CHAPTER XI 

ORDINARY DIFFERENTIAL EQUATIONS WITH THREE 
VARIABLES, AND THE CORRESPONDING CURVES AND 
SURFACES 

111. We shall now consider some simple differential equations 
expressing properties of curves in space and of surfaces on which 
these curves lie, or which they cut orthogonally (as in Electro- 
statics the Equipotential Surfaces cut the Lines of Force ortho- 
gonally). The ordinary * differential equations of this chapter are 
closely connected with the partial differential equations of the 
next. 

Before proceeding further the student should revise his solid 
geometry. We need in particular the fact that the direction-cosines 
of the tangent to a curve are 



(dx dy dz\ 
\ds' ds' ds/' 



i.e. are in the ratio dx:dy:dz. 

Simultaneous linear equations with constant coefficients have 
already been discussed in Chapter III. 

112. The simultaneous equations — r = rT = p- These equations 

express that the tangent to a certain curve at any point (x, y, z) 
has direction-cosines proportional to (P, Q, R). If P, Q, and R are 
constants, we thus get a straight line, or rather a doubly infinite 
system of straight lines, as one such line goes through any point of 
space. If, however, P, Q, and R are functions of x, y, and z, we get 
a similar system of curves, any one of which may be considered as 
generated by a moving point which continuously alters its direction 

* i.e. not involving partial differential cofficients. 
133 



J 



1/ 



134 DIFFERENTIAL EQUATIONS 

of motion. The Lines of Force of Electrostatics form such a 
system.* 

Ex. (i). tejlji (!) 

Obvious integrals are x-z = a, (2) 

y-z = b, (3) 

the equations of two planes, in tersecting in the line 

x-a y-b 2 

— -7T-V (4) 

which by suitable choice of the arbitrary constants a and b can be made 
to go through any given point, e.g. through (/, g, h) if a=f-h and 
b =g - h. 

Instead of picking out the single line of the system that goes through 
one given point, we may take the infinity of such lines that intersect 
a given curve, e.g. the circle x 2 + y 2 = i, 2=0. 

The equations of this circle, taken together with (2) and (3), give 

x = a, 

y=b, 

and hence a z + b 2 = i (5) 

This is the relation that holds between a and b if the line is to inter- 
sect the circle. Eliminating a and b from (2), (3), and (5), we get 

(x-z) 2 + (y-z) 2 = 4, 

the elliptic cylinder formed by those lines of the system which meet 
the circle. 

Similarly the lines of the system which meet the curve 

<p(x,y)=:0, 2 = 

form the surface <p(x-z, y - z) = 0. 

th /--x dx dy dz ,_, 

Ex -<">- 7-0 -~x (6) 

Obvious integrals are x 2 + z 2 = a, (7) 

y=b, (8) 

a right circular cylinder and a plane that cuts it in a circle. 

The differential equations therefore represent a system of circles, 
whose centres all lie on the axis of y and whose planes are all perpen- 
dicular to this axis. 

One such circle goes through any point of space. That through 
(/> 9s h ) is x 2 + z 2 =/ 2 + h 2 , y =g. 

A surface is formed by the circles of the system that intersect a 
given curve. 

* The equations of the lines of force are dx ^~—dy /W- = rfz 'r>T> wnere 
V is the potential function. I & I <% ' Cz 



ORDINARY EQUATIONS WITH THREE VARIABLE 135 

If the given curve is the hyperbola 

£!_l_ 2 _i z - 

42 B 2 ~ 

(7) and (8) give, for a circle intersecting this hyperbola, 

x 2 = a, y = b, 

and hence T" 2_ fi" 2 = 1 ^ 

Eliminating a and b from (7), (8), and (9), we get the hyperboloid 
of one sheet, x 2 + z 2 ^2 

~A 2 fii" 1, 

formed by those circles of the system that intersect the hyperbola. 

Similarly, starting from the curve <p(x 2 , y)=0, 2 = 0, we get the 
surface of revolution (p{x 2 + z 2 , ?/)=0. 

113. Solution of such equations by multipliers. If 

dx dy dz 

each of these fractions is equal to 

Idx + mdy + ndz . ' f 

IP+mQ + nR ' ." * 

This method may be used with advantage in some examples to 
obtain a zero denominator and a numerator that is an exact 
differential. {[My^,'^' S 

Ex . to dy_ te / L^ 

z(x + y) z(x-y) x 2 + y* / 

Each of these fractions 

x dx - y dy - z dz 



xz(x + y) -yz(x-y) -z(x 2 + y 2 ) 
xdx-ydy-zdz 

= ; 

therefore x dx - y dy - z dz = 0, 

i.e. x 2 -y 2 -z 2 = a (2) 

„. ., , , . . ydx + xdy — zdz 
(Similarly, each fraction = - pp ', 

therefore ydx + xdy-zdz = 0, 

i.e. 2xy-z 2 = b (3) 

Thus the solution of equations (1) is formed by the system of quartic 
curves in space arising from the intersection of the conicoids (2) and 
(3), where a and b are arbitrary. 



136* Differential equations 

Examples for solution. 

Obtain the system of curves, defined by two equations with an 
arbitrary constant in each, satisfying the following simultaneous dif- 
ferential equations. Interpret geometrically whenever possible. 

>i a) ^ = ^/ = ^. », ( 2 ) dx = d y _ dz 

x y z' \s mz-ny nx-lz ly-mx' 

dx dy dz \1a\ dx _dy _dz 

y 2 + z 2 -x 2 -2xy -2xz' yz' zx xy' 

dx _ dy _ dz xdx dy _ dz 

y + z z + x x + y' z 2 -2yz-y 2 y + z y-z' 

(7) Find the radius of the circle of Ex. 2 that goes through the 
point (0, -n, m). 

(8) Find the surface generated by the curves of Ex. 4 that intersect 
the circle y 2 + z 2 = \, x=0. 

(9) Find the surface generated by the lines of Ex. 1 that intersect 

the helix x 2 + y 2 = r 2 , z = &tan -1 -- 
y x 

(10) Find the curve which passes through the point (1, 2, -1) and 
is such that at any point the direction-cosines of its tangent are in the 
ratio of the squares of the co-ordinates of that point. 

114. A second integral found by the help of the first. Consider the 
equations dx dy _ dz m 

T ~~^2 "3a; 2 sin (y +2x) * ' 

An obvious integral is y +2x=a (2) 

Using this relation, we get 

dx _ dz 
1 3a; 2 sin a 

giving z-x 3 sin a = b. 

Substituting for a, z - x 3 sin (y + 2x) = b (3) 

Is (3) really an integral of (1) ? 

Differentiating (3), 

{dz - 3x 2 dx sin (y + 2a;)} - x 3 cos (y + 2x) . {dy + 2 dx} =0, 
which is true in virtue of (1). So (3) is an integral. 

Examples for solution. 

\j m dx _ d y_ dz V (2) dx _ d y _ dz 

V K ' 1 3 5z + tan(?/-3a;)' K ' z -z z 2 + (y + x) 2 ' 

/ (3) dx = dy ^dz dx _dy _ dz 

^ xz(z 2 + xy) -yz(z 2 + xy) x 4 ' ^ xy y 2 zxy-2x 2 ' 



ORDINARY EQUATIONS WITH THREE VARIABLES 137 

115. General and special integrals of simultaneous equations. If 
w=aandu=6 are two independent integrals of the simultaneous 
equations dx dy dz 

P~Q~R' 

then <p(u, v)=0 represents a surface passing through the curves of 
the system, and should therefore give another solution, whatever 
the form of the function (p. 

An analytical proof of this is reserved for the next chapter, as 
its importance belongs chiefly to partial differential equations. 

(j)(u, v)=0 is called the General Integral. Some simultaneous 
equations possess integrals called Special, which are not included in 
the General Integral. 

Examples for solution. 

(1) In the Ex. of Art. 113 u = x 2 -y 2 -z 2 and v = 2xy-z 2 , so the 
General Integral is <p(x 2 -y 2 -z 2 , 2xy-z 2 )=0. The student should 
verify this in the simple cases where 

<b(u. v)=u-v or ik(u, v)= -. 

r • u-2 

(2) Verify that for the equation 

dx dy dz 

l+^/(z-x-y) = ~l = ~2' 
the General Integral may be taken as 

</>{2y -z,y + 2y/{z -x-y)} =0, 
while z = x + y is a Special Integral. 

116. Geometrical interpretation of the equation 

Pdx+Qdy+Rdz = 0. 

This differential equation expresses that the tangent to a curve 
is perpendicular t« a certain line, the direction-cosines of this tangent 
and line being proportional to (dx, dy, dz) and (P, Q, R) respectively. 
But we saw that the simultaneous equations 
dx _dy _dz 
P = Q = R 
expressed that the tangent to a curve was parallel to the line (P, Q, R). 
We thus get two sets of curves. If two curves, one of each set, 
intersect, they must intersect at right angles. 

Now two cases arise. It may happen that the equation 
Pdx+Qdy+Rdz=0 
is integrable. This means that a family of surfaces can be found, 
all curves on which are perpendicular to the curves represented by 



138 DIFFERENTIAL EQUATIONS 

the simultaneous equations at all points where these curves cut the 
surface. In fact, this is the case where an infinite number of surfaces 
can be drawn to cut orthogonally a doubly infinite set of curves, 
as equipotential surfaces cut lines of force in electrostatics. On the 
other hand, the curves represented by the simultaneous equations 
may not admit of such a family of orthogonal surfaces. In this 
case the single equation is non-integrable. 

Ex. (i) . The equation dx + dy + dz=0 
integrates to x + y + z = c, 

a family of parallel planes. 

We saw in Ex. (i) of Art. 112 that the simultaneous equations 
dx dy dz 

represented the family of parallel lines 

x - a y-b z 

The planes are the orthogonal trajectories of the lines. 

Ex. (ii). zdx-xdz=0, 

dx dz - 
i.e. = 

x z 

integrates to z — cx, 

a family of planes passing through the axis of y. 

We saw in Ex. (ii) of Art. 112 that the corresponding simultaneous 
equations dxjj^dz^ 

z -x 

represented a system of circles whose axes all lie along the axis of y, 
so the planes are the orthogonal trajectories of the circles. 

Examples for solution. 

Integrate the following equations, and whenever possible interpret 
the results geometrically and verify that the surfaces are the orthogonal 
trajectories of the curves represented by the corresponding simultaneous 
( equations : 

V (1) xdx + ydy + zdz = 0. 

(2) (y 2 + z 2 - x 2 ) dx - 2xy dy - 2xz dz = 0. [Divide by x 2 . ] 
V(3) yzdx + zxdy + xydz=0. V(4) (y + z)dx + (z + x)dy + (x + y)dz=0. 

V (5) z(ydx-xdy)—y 2 dz. ]/ (6) xdx + zdy + (y + 2z)dz=0. 

117. Method of integration when the solution is not obvious. When 
an integrable equation of the form 

Pdx+Qdy + Rdz=0 




ORDINARY EQUATIONS WITH THREE VARIABLES 139 

cannot be solved by inspection, we seek for a solution by considering 
first the simpler case wh ere z is constant and so dz=0* 

For example, yzdx + 2zxdy -3xydz=0 becomes, if z is constant, 
ydx+2xdy=0, 
giving xy 2 = a. 

As this was obtained by supposing the variable z to be constant, 
it is probable that the solution of the original equation can be 
obtained by replacing the constant a by some function of z, giving 

leading to y 2 dx + 2xy dy - J- <fo*= 0. 

This is identical with the original equation if 

Jf 
y z _ 2xy 

yz 2zx 

df _3xy 2 

dz z 

df_3dz 

7" * J 

f(z)=cz 3 , 
giving the final solution xy 2 = cz 3 . 

For a proof that this method holds good for\all integrable.j 
e quations, see Art. 119. 

Examples for solution. 
ls(l) yz log zdx-zx log zdy + xy dz=0. 
1/(2) 2yz dx + zxdy-xy(l + z)dz = 0. 

(3) (2x 2 + 2xy + 2xz 2 + l)dx + dy + 2zdz = 0. [N. B— Assume x con- 
stant at first. ] 

(4) (y 2 + yz) dx + (zx + z 2 ) dy + (y 2 - xy) dz = 0. 

/- (5) (x 2 y -y 3 - y 2 z) dx + (xy 2 - x 2 z - x 3 ) dy + (xy 2 + x 2 y) dz=0. 

(6) Show that the integral of the following equation represents a 
family of planes with a common line of intersection, and that these 
planes are the orthogonal trajectories of the circles of Ex. 2 of the set 
following Art. 113 : 

(mz - ny) dx + (nx - Iz) dy + (ty- nix) dz ■— 0. 

118. Condition necessary for an equation to be integrable. If 

Pdx+Qdy+Rdz=0 (1) 

has an integral <f)(x, y, z) =c, which on differentiation gives 

~ dx + ^ dy + -_- dz - 0, 
ox dy J dz 



140 DIFFERENTIAL EQUATIONS .-, 

*» «*-XP; j*-X«; g-XS. 

^(f-fHI-*|-° < 2 ) 

*(?-©♦*£-«£- « 

Multiply equations (2), (3), and (4) by P, #, and R respectively, 
and add. We get 

^-|)+<-!)+<-!)=o. 

If the equation (1) is integrable, this condition must be satisfied. 
The student familiar with vector analysis will see that if P, Q, R 
are the components of a vector A, the condition may be written 

A .curl A=0. 
Ex. In the worked example of the last article, 
yz dx + 2zx dy - 3xy dz = 0, 
P = y z > Q = 2z#, R = - 3xy. 
The condition gives 

yz (2x + 3x) +2zx(- 3y -y)- 3xy (z - 2z) = 0, 
i.e. 5xyz - 8xyz + 3xyz = 0, 
which is true. 

Examples for solution. 

(1) Show that the equations in the last two sets of examples 
satisfy this condition. 

(2) Show that there is no set of surfaces orthogonal to the curves 
given by dx__dy dz 

z x + y 1 

* 119. The condition of integrability is sufficient as well as necessary. 
We shall prove that the condition is sufficient by showing that 
when it is satisfied the method of Art. 1 17 will always be successful 
in giving a solution. 

We require as a lemma the fact that if P, Q, R satisfy the con- 
dition, so also do P 1 = XP, Q x -=XQ, R X =XR, where X is any function 
of x, y, and z. We leave this as an exercise to the student. 

* To be omitted on a first reading. 



ORDINARY EQUATIONS WITH THREE VARIABLES 141 

In Art. 117 we supposed a solution of 
Pdx+Qdy=0 
obtained by considering z as constant. 
Let this solution be F(x, y, z) =a, 

which gives -=- dx+-~- dy=0, 

dFlj, dFl n ^ 

Put \P = P V \Q = Qi, XR = R 1 . 

The next step was to replace a by f(z), giving 

F(x,y,z)=f(z), (1) 

j a i. Ms dF i i dF d f\s a 

and thence v— dx + ~- dy + 1 » — j- Ydz=0, 

i.e. p i dx+Q 1 dy + \^-^ ]dz=0 (2) 

This is identical with 

Pdx+Qdy + Rdz=0, 

« if I-"-* < 3 > 

In the example of Art. 117 we got 

dfjxy*_3f(z) 

dz z z ' 

the a; and y being got rid of by virtue of the equation x 2 y =f(z). 

What we have to prove is that the x and y can always be got rid 

of from the right-hand side of equation (3) in virtue of equation (1). 

dF 
In other words, we must show that -R, involves x and y 

dz 
only as a function of F. 

Now this will be the case if * 

dFd IdF „\ dFd idF .,) .., ,. „ ... 

dxdyUz -M -Tydx[dz "**} = ° ldentlCaU " V (4) 

Now, by the lemma, the relation between P, Q, R leads to the 
similar relation 

\dQ x dR,\ idR, dP,) fdP t BQ,} 

F n'dz'^yi +Ql Vdx~'dz~j +Kl \dJ,~dx~j ~°' 

* Edwards' Differential Calculus, Art. 510. 



142 DIFFERENTIAL EQUATIONS 

also, since equation (2) is integrable, 



x \dz dy\dz dz)i +Vl \dx\dz TzJ"Wj 

\oz dz/ [dy dx) 
By subtraction of these last two equations we get 

Pl By\dz~ ~dz~ Rl ) ~ Ql dx[dz 'Jz'^f 

-{M-«.}{f-iH ,) 

But .,,.£, Ql Jl, and |(|)=i(|)-0, 

as / is a function of z alone. 
Hence (5) reduces to (4). 

That is, ^ — J? x can be expressed as a function of F and z, say 
\},(F, z). Hence from (1) and (3), 

If the solution of this is f= x (z), then i^(x, ?/, z)=x(z) is a 
solution of Pdx+Qdy+Rdz=0, 

which is thus proved to be integrable whenever P, Q, R satisfy the 
condition of Art. 118. 

120. The non-integrable single equation. When the condition of 
integrability is not satisfied, the equation 

Pdx+Qdy+Rdz=0 (1) 

represents a family of curves orthogonal to the family represented 
by the simultaneous equations 

dx dy dz 
P = Q = R' 
but in this case there is no family of surfaces orthogonal to the 
second family of curves. 

However, we can find an infinite number of curves that lie on 
any given surface and satisfy (1), whether that equation is integrable 
or not. 

Ex. Find the curves represented by the solution of 

y dx + (z - y) dy + x dz =0, (1) 

which lie in the plane 2x-y-z = ] (2) 

(It is easily verified that the condition of integrability is not satisfied.) 



ORDINARY EQUATIONS WITH THREE VARIABLES 143 

The method of procedure is to eliminate one of the variables and 
its differential, say z and dz, from these two equations and the differ- 
ential of the second of them. 

Differentiating (2), 2dx -dy-dz =0. 

Multiplying by x and adding to (1), 

(y + 2x)dx + (z-x- y) dy = 0, 
or using (2), (y + 2x) dx + (x - 2y - 1 ) dy = 0, 
which gives xy + x 2 -y 2 -y = c 2 (3) 

Thus the curves of the family that lie in the plane (2) are the sections 
by that plane of the infinite set of rectangular hyperbolic cylinders (3). 

The result of this example could have been expressed by saying 
that the projections on the plane of xy of curves which lie in the plane 
(2) and satisfy equation (1) are a family of concentric, similar and 
similarly situated rectangular hyperbolas. 

Examples for solution. 

(1) Show that there is no single integral of dz = 2y dx + x dy. 
Prove that curves of this equation that lie in the plane z = x + y lie 

also on surfaces of the family (x - l) 2 {2y - 1) =c. 

(2) Show that the curves of 

// x v \ 
xdx + ydy + cyj\\ — a~ jb j dz=0 

that lie on the ellipsoid 

x 2 y 2 z 2 , 
— +— + — = 1 
a 2 b 2 c 2 

lie also on the family of concentric spheres 

x 2 + y 2 + z 2 = k 2 . 

(3) Find the orthogonal projection on the plane of xz of curves 
which lie on the paraboloid 3z=x 2 + y 2 and satisfy the equation 

2dz = {x + z) dx + y dy. 

(4) Find the equation of the cylinder, with generators parallel to 
the axis of ?/, passing through the point (2, 1, - 1), and also through a 
curve that lies on the sphere x 2 + y 2 + z 2 = 4: and satisfies the equation 

(xy + 2xz) dx -f y 2 dy + (a: 2 + yz) dz = 0. 

MISCELLANEOUS EXAMPLES ON CHAPTER XI. 

,„ v dx dy dz £ni dx dy dz 

xz yz xy' y 3 x-2x i 2y /l -x z y ( dz(x' A -if)' 

... dy dz 

dx dy dz 

(4) (z + z 3 ) cosxj t ~{z + z 3 ) J + (1 - z 2 ) (y - sin r) ~ = 0. 



144 DIFFERENTIAL EQUATIONS 

(5) (l. + ,. + l«)$ + *» J+-J-1. 

(6) Find f(y)iff (y) dx -zxdy- xy log y dz = is integrable. 
Find the corresponding integral. 

(7) Show that the following equation is not integrable : 

3y dx + (z- 3y) dy + xdz=0. 
Prove that the projection on the plane of xy of the curves that 
satisfy the equation and lie in the plane 2x + y - z = a are the rectangular 
hyperbolas x 2 + 3xy-y 2 -ay = b. 

(8) Find the differential equations of the family of twisted cubic 
curves y = ax 2 ; y 2 = bzx. Show that all these curves cut orthogonally 
the family of ellipsoids 

x 2 + 2y 2 + 3z 2 = c 2 . 

(9) Find the equations of the curve that passes through the point 
(3, 2, 1) and cuts orthogonally the family of surfaces x + yz = c. 

(10) Solve the following homogeneous equations by putting x = uz, 

y = vz : 

(i) (x 2 -y 2 -z 2 + 2xy + 2xz) dx + (y 2 -z 2 -x 2 + 2yz + 2yx) dy 

+ (z 2 -x 2 -y 2 + 2zx + 2zy)dz=0; 
(ii) (2xz - yz) dx + (2yz - xz) dy - (x 2 -xy + y 2 ) dz = Q; 
(iii) z 2 dx + (z 2 - 2yz) dy + (2y 2 -yz- xz) dz=0. 

(11) Prove that if the equation 

P 1 dx l + P 2 dx 2 + P 3 dx 3 + P^dx^ = 
is integrable, then 

\ox t ox s J \ax r ox t / \ox s ox r / 
where r, s, I are any three of the four suffixes 1, 2, 3, 4. 
Denoting this relation by C rst = 0, verify that 

^1^234 - ^2^134 + ^3^124 - ^A 23 = identically, 
showing that only three of these four relations are independent. 
Verify that these conditions are satisfied for the equation 

+ (x 3 2 - x 1 x 2 x i ) dx 3 + (# 4 2 - x 1 x 2 x 3 ) dx A =0. 

(12) Integrate the equation of Ex. 11 by the following process: 
(i) Suppose x 3 and x A constant, and thus obtain 

X-t ~r Xn Ti/ziXaX*^ == CI, 

(ii) Replace a by f (x 3 , x A ). By differentiation and comparison with 

the original equation obtain -= -, =— , and hence /"and the solution 
' ox 3 dx 4 J 



MISCELLANEOUS EXAMPLES 145 

(13) Integrate the equation of Ex. 11 by putting x 1 =ux 4 , x 2 = vx 4 , 
\x 3 = ivx A . 

(14) Show that the following equation satisfies the conditions of 
integrability and obtain its integral : 

y sin wdx + x sin w dy - xy sin wdz-xy cos w dw = 0. 

(15) Show that the equation 

adx 2 + bdy 2 + cdz 2 + 2fdydz + 2gdzdx + 2hdxdy = 
reduces to two equations of the form 

Pdx + Qdy+Rdz = 
if <tic + 2fgh - af 2 - bg 2 - ch 2 =0. (Cf . a result in Conies.) 

Hence show that the solution of 
xyz (dx 2 + dy 2 + dz 2 ) +x(y 2 + z 2 ) dydz + y {z 2 + x 2 ) dz dx 

+ z(x 2 + y 2 ) dxdy = 
is (x 2 + y* + z z-c) {xyz - c) =0. (Cf. Art. 52.) 

(16) Show that the condition of integrability of 

Pdx+Qdy + Rdz = (1) 

implies the orthogonality of any pair of intersecting curves of the 
families 

dx/P = dy/Q = dz/R (2) 

- *ra-s-*/«-©.-*/s-g) '*> 

Hence show that the curves of (3) all lie on the surfaces of (1). 
Verify this conclusion for P=ny-mz, Q = lz-nx, R-mx-ly. 
(For the solutions of the corresponding equations, see earlier examples 
in this chapter.) 

(17) The preceding example suggests that if a = const., /3 = const. 
are two integrals of equations (3), the integral of equation (1) should 
be expressible in the form /(a, /3)=const., and hence that 

P dx + Q dy + R dz 
should be expressible as Adu + B dj3, where A and B are functions of 
- and ft. 

Verify that for the case 

P = yz log z, Q= -zx log z, R = xy, 
a = yz-, ft = xz*\ogz, A=-ft, and B = a. 
Hence obtain an integral of (1) in the form a = cft, 
i.e. y=cx\ogz. 



CHAPTER XII 

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST 
ORDER. PARTICULAR METHODS 

121. We have already (in Chap. IV.) discussed the formation of 
partial differential equations by elimination of arbitrary functions 
or of arbitrary constants. We also showed how in certain equations, 
of great importance in mathematical physics, simple particular 
solutions could be found by the aid of which more complex solutions 
could be built up to satisfy such initial and boundary conditions as 
usually occur in physical problems. 

In the present chapter we shall be concerned chiefly with equa- 
tions of geometrical interest, and seek for integrals of various forms, 
" general," " complete," and " singular," and their geometrical 
interpretations. Exceptional equations will be found to possess 
integrals of another form called " special." 

122. Geometrical theorems required. The student should revise 
the following theorems in any treatise on solid geometry : 

(i) The direction-cosines of the normal to a surface f (x, y, 2) =0 
at the point (x, y, z) are in the ratio 

dx ' dy ' dz * 
Since 

dfjdf dz , dfldf dz 

dxjdz=dx=¥>™T> and - d yli- d -y^>™y> 

this ratio can also be written p : q : - 1 . 

The symbols p and q are to be understood as here defined all 
through this chapter. 

(ii) The envelope of the system of surfaces 
f(x,y,z,a,b)=0, 



PARTICULAR METHODS 147 

where a and b are variable parameters, is found by eliminating 
a and b from the given equation and 

da u ' db U * 

The result may contain other loci besides the envelope (cf. 
Chap. VI.). 

123. Lagrange's linear equation and its geometrical interpretation. 
This is the name applied to the equation 

Pp+Qq=R, (1) 

where P, Q, R are functions of x, y, z. 

The geometrical interpretation is that the normal to a certain 
surface is perpendicular to a line whose direction-cosines are in the 
ratio P : Q : R. But in the last chapter we saw that the simultaneous 
equations dx _dy _dz 

P'Q-R ( } 

represented a family of curves such that the tangent at any point 
had direction-cosines in the ratio P :Q: R, and that <p (u, v)=0 
(where u= const, and v= const, were two particular integrals of 
the simultaneous equations) represented a surface through such 
curves. 

Through every point of such a surface passes a curve of the 
family, lying wholly on the surface. Hence the normal to the 
surface must be perpendicular to the tangent to this curve, i.e. 
perpendicular to a line whose direction-cosines are in the ratio 
P : Q : R. This is just what is required by the partial differential 
equation. 

Thus equations (1) and (2) are equivalent, for they define the 
same set of surfaces. When equation (1) is given, equations (2) are 
called the subsidiary equations. 

Thus cp (u, v)=0 is an integral of (1), if u = const, and v = const, 
are any two independent solutions of the subsidiary equations (2) 
and cf> is any arbitrary function. This is called the General Integral 
of Lagrange's Linear Equation. 

Ex. (i). p + q = l. 

The subsidiary equations are those discussed in Ex. (i) of Art. 112, 
viz. dx dy dz 

T = T = T' 

representing a family of parallel straight lines. 



1 



148 DIFFERENTIAL EQUATIONS 

Two independent integrals are 

x-z-a t 
y-z = b, 
representing two families of planes containing these straight lines. 

The general integral is <p(x-z, y-z)=0, representing the surface 
formed by lines of the family passing through the curve 

(p(x,y)=0, 2=0. 
If we are given a definite ciirve, such as the circle 
x 2 + y 2 = i, z*=0, 
we can construct a corresponding particular integral 

(x -z)* + (y-z)* = l, 
the elliptic cylinder formed by lines of the family meeting the given 
circle. 

Ex. (ii). zp = -x. [Cf. Ex. (ii) of Art. 112.] 
The subsidiary equations are 

dx cly dz 
2 x ' 

of which two integrals are x 2 + z 2 = a, y = b. 

The general integral <p(x 2 + z 2 , y)=0 represents the surface of 
revolution formed by curves (circles in this case) of the family inter- 
secting the curve ( x * } y ) = o, 2=0. 

Ex. (iii). Find the surfaces whose tangent planes cut off an intercept 
of constant length k from the axis of z. 
The tangent plane at (x, y, z) is 

Z-z=p(X-x)+q(Y-y). 
Putting A' = Y = 0, Z = z-px- qy = k. 
The subsidiary equations are 

dx dy dz 
x y z-k 
of which y — ax, z-k — bx, are integrals. 

The general integral <p(~, - —J=0 represents any cone with it3 

vertex at (0, 0, k), and these surfaces clearly possess the desired property. 

Examples for solution. 

Obtain general integrals of the following equations. [Cf. the first 
set of examples in Chap. XL] 

(1) xp + yq = z. 

(2) (»iz - ny) p + (».r - Iz <} = ly- mx. 

(3) {if + z 2 - x 2 ) j> - 2xy g + 2xz= 0. 

(4) yzp + zxq = xy. 

(5) (y + z)p + (z + x)q = x + y. 



PARTICULAR METHODS 149 

(6) (z 2 - 2yz -y 2 )p + (xy + xz)q = xy - xz. 

(7) p + 3q = 5z + tan(y-3x). 

(8) zp-zq = z 2 + (y + x) 2 . 

(9) Find a solution of Ex. (1) representing a surface meeting the 
parabola y 2 = 4:X, 2 = 1. 

(10) Find the most general solution of Ex. (4) representing a conicoid. 

(11) Show that if the solution of Ex. (6) represents a sphere, the 
centre is at the origin. 

(12) Find the surfaces all of whose normals intersect the axis of z. 

124. Analytical verification of the general integral. We shall now 
eliminate the arbitrary function <p from <p(u, v)=0, and thus 
verify analytically that this satisfies Pp +Qq=R, provided u = a and 
v = b are two independent* integrals of the subsidiary equations 

dx __dy _dz 
P~Q~R' 
Differentiate (u, v)=0 partially with respect to x, keeping y 
constant ; z will vary in consequence of the variation of x. Hence 

we get fyf 9 ^ du <k\ dj>(dv dv dz\ 

du\dx dz dx) dv^dx dzdx) ' 

9r/> (du du\ dcp /dv dv 

~du\dx dz) dv\dx ^ dz. 

. ., , drh/du du\ d(h/dv dv\ _ 

Similarly £(^ + f s ) + £ (^ + , jj - 0. 

Eliminating the ratio -zj- : ~ from these last two equations, 

(du du\ (dv dv\ fdu du\ (dv dv 
\Fy +q Tz){dx + ^Tz)^\¥x + ^^J\dlj + (I dz 
/du dv du dv\ /du dv du dv\ 
\dydz dzdy) * \dzdx dxdz) * 
_ du dv du dv 
dx dy dydx 

But from u =a, -_ dx + _ dy + ' dz = 0, 
dx dy J dz 

and hence from the subsidiary equations, of which u =a is an integral, 

„ du ~du „ du . 
Pw-+Q-^-+R- a - =0. 

dx dy oz 

*If u and v are not independent, (^^i -^^\ and the other two similar 

\ cy dz dz ay I 
expressions all vanish identically (Edwards' Differential Calculus, Art. 510), which 
reduces equation (1) to 0=0 



(1) 



150 DIFFERENTIAL EQUATIONS 

pg + eg + *|=o. 

Hence 

P n 7? - /^ u ^ v ^ u ^ v \ • ^ M ^ v ^ u ^ v \ • ^" ^ v ^ u ^ v \ • 

' ' \dy dz dz dy) \dzdx dx dzJ ' \dx dy By dx/ ' 

so (1) becomes Pp + Qq=R, the equation required. 

125. Special integrals. It is sometimes stated that all integrals 
of Lagrange's linear equation are included in the general integral 
<p (u, v) = 0. But this is not so. 

For instance, the equation 

p-q = 2y / z 
has as subsidiary equations 

dx _ dy _ dz 
T = ^l _ 2Vz* 

Thus we may take u=x+y,v=x- s/z, and the general integral as 
<p(x+y, x- <y/z)=Q. 

But 2=0 satisfies the partial differential equation, though it is 
obviously impossible to express it as a function of u and v. 

Such an integral is called special. It will be noticed that in all 
the examples given below the special integrals occur in equations 
involving a term which cannot be expanded in series of positive 
integral powers. 

In a recent paper M. J. M. Hill* has shown that in every case 
where special integrals exist they can be obtained by applying a 
suitable method of integration to the Lagrangian system of sub- 
sidiary equations (see Examples 5 and 6 below). He also under- 
takes the re-classification of the integrals, the necessity of which 
task had been pointed out by Forsyth. f 

Examples for solution. 

Show that the following equations possess the given general and 
special integrals : 



(1) 


V 


+ 2qz h = 3; 


s ; </>(•£- 


z-\ y-z* 


= 


> ~ 


= 0. 








(2) 


V 


+ q{l+(z- 


-y)*}=i; 


(j){x-Z, 


2x + 3(c 


-y) 1 }; 


z = 


y- 




(3) {1 


+ y/(z-x 

z = x + y. 


-y)}p+<i 


= 2; <p{ 


2//- 


z, y 


f2V(* 


-X 


-y)}=0; 

[Chrj 


•stal.] 








* Proc. Lot 


ulon Math. 


Soc. 


1017 
















t Proc. Lo> 


don Math. 


Soc. 


1905 


0. 









PARTICULAR METHODS 151 

(4) By putting (z-x-y)* = w in Chrystal's equation (Ex. 3), obtain 






dy 
This shows that z-x-y=0 is a solution of the original equation. 

[Hill.] 

(5) Show that the Lagrangian subsidiary equations of Chrystal's 
equation (Ex. 3) may be written 

dx , . A dz „ 

and deduce that j-{z-x-y) = -(z-x- y) , 

of which z - x - y = is a particular solution. [Hill.] 

(6) Obtain the general and special integrals of the equation 

p-q = 2\/z 
by imitating Hill's methods as given in Exs. 4 and 5. 

126. The linear equation with n independent variables. The 

general integral of the equation 

PlPl + P 2P* + P&3 + ■■■+ PnVn = R, 

where p 1 =^-, p 2 =^—, ••• etc., and the P's and R are functions 

of the x's and z, is <p(u 1} w 2 < u 3 , ... u n ) = 0, 

where % = const., u 2 = const., ... etc., are any n independent integrals 

of the subsidiary equations 

ttwi tviX'o U/Juo iX&t 

This may be verified as in Art. 124. The student should write 
out the proof for the case of three independent variables. 

Besides this general integral, special integrals exist for excep- 
tional equations, just as in the case of two independent variables. 

Examples for solution. 

(1) P2 + P3 = 1 +PV 

(2) x 1 p l + 2x 2 p 2 + 3x 3 p 3 + 4 : x i p 4 = Q. 

(3) (x 3 - x 2 )p x + x 2 p 2 - x 3 p 3 = x 2 (x l + x 3 ) - x 2 2 . 

(4) x 2 x 3 p x + x 3 x 1 p 2 + x l x 2 p 3 + x 1 x 2 x 3 =0. 

(5) p 1 + x 1 p 2 + x 1 x 2 p 3 = x l x 2 x 3 \ / z. 

(6) p x +p 2 + p 3 {l + \/{z - *i - J' 2 - a 3 )} = 3. 



152 DIFFERENTIAL EQUATIONS 

Pf 7W "pic 

127. The equation P =- +Q =- +R =- -0. If P, Q, R are functions 
ox dy oz 

of x, y, z but not of /, the equation can be viewed from two different 

aspects. 

Consider, for example, 

I-g + V*I=o (i) 

We may regard this as equivalent to the three-dimensional 

equation p-q = 2y/z, (2) 

of which <p(x+y, x- \/z)=0 is the general integral and 2=0 a 
special integral. 

On the other hand, regarding (1) as an equation in four variables, 
we get the general integral 

<t>(f>to+V, x-\/z)=0, 
which is equivalent to f=\fs{x +y, x- y/z), where \[s is an arbitrary 
function, but if 

Thus/ = z is not an integral of (1), although f = z=0 certainly 
gives a solution. 

In general it may be proved that 

regarded as four-dimensional, where P, Q, R do not contain/, has 
no special integrals.* A similar theorem is true for any number of 
independent variables. 

Examples for solution. 

(1) Verify that if f=x, f=0 is a surface satisfying 

V*Iw y g + V*f=o, 

and hence that this differential equation, interpreted three-dimension- 
ally, admits the three special integrals x=0, y = 0, z=Q and the general 
integral <j>(\/z- \/x, y/z - y/y) = 0. 

(2) Show that the general integral of the last example represents 
surfaces through curves which, if they do not go through the origin, 
either touch the co-ordinate planes or lie wholly in one of them. 

[Hint. Prove that -r = \l\ ), and that dr,/ds = if x=0, 

ds \ \x + y + zJ ' 

unless x, y, z are all zero.] 

* See Appendix B. 






PARTICULAR METHODS 153 

(3) Show that s/x - + y/y «- =0, regarded two-dimensionally, repre- 
sents a family of parabolas \/y = yjx + c, and their envelope, the 
co-ordinate axes x*=Q, y=0; while regarded three-dimensionally it 

represents the surfaces z = <p(y* ~ar). 

128. Non-linear equations. We shall now consider equations in 
which p and q occur other than in the first degree. Before giving 
the general method we shall discuss four simple standard forms, for 
which a " complete integral " (i.e. one involving two arbitrary 
constants) can be obtained by inspection or by other simple means. 
In Arts. 133-13.5 we shall show how to deduce general and singular 
integrals from the complete integrals. 

129. Standard I. Only p and q present. Consider, for example, 
this equation q =3jo 2 . 

The most obvious solution is to take p and q as constants satisfying 
the equation, say p = a, q— 3a 2 . 

Then, since dz=pdx +q dy = adx+3a 2 dy, 

z = ax + 3a 2 y + c. 
This is the complete integral, containing two arbitrary constants 
a and c. 

In general, the complete integral oif(p, q) = is 
z = ax+by + c, 
where a and b are connected by the relation /(a, 6) =0. 

Examples for solution. 

Find complete integrals of the following : 
(1) p = 2q 2 + l. (2) p 2 + q 2 = l. 

(3) p = e*. (4) pY = l. 

(5) 2?2- 9 2 = 4. (6) pq = v + q. 

130. Standard II. Only p, q, and z present. Consider the equation 

z 2 (ph 2 +q 2 )=l (1) 

As a trial solution assume that z is a function of x + ay 
( =u, say), where a is an arbitrary constant. 

' , dz _dz du _dz _dz _dz du _ dz 

dx du dx du' dy du dy du' 

/dz\ 2 

Substituting in (1), z 2 (~) ^ + r{2 ) = 1 ' 

du „ £ 

i.e. -j- = ±Z .- +a 2 ) , 
dz 

i.e. u + b = ±l (z 2 + a 2 Y, 
i.e. 9(x +ay +b) 2 = (z 2 +a 2 ) 3 . 



154 DIFFERENTIAL EQUATIONS 

In general, this method reduces f(z,p, q)=0 to the ordinary 
differential equation 



/(*> 



dz dz\_~ 
du ' duJ 



Examples for solution. 

Find complete integrals of the following : 
(1) iz = pq. (2) z 2 = l+p 2 + q 2 . 

(3) q 2 = z 2 p 2 (\ -p 2 ). (4) f + q A = 21z. 

(5) p(z + p)+q = 0. (6) p 2 = zq. 

L31. Standard III. f(x, p)=F(y, q). Consider the equation 

p -3x 2 =q 2 -y. 
As a trial solution put each side of this equation equal to an 
arbitrary constant a, giving 

p=3x 2 +a; q = \ / (y +a). 
But dz=pdx+qdy 

- (3x 2 +a)dx + \/(y +a)dy ; 
t h er ef or e z = x 3 + ax + § (y + a)'- + b, 

which is the complete integral required. 

Examples for solution. 

Find complete integrals of the following : 

(1) p 2 = q + x. (2) pq=xy. 

(3) yp = 2yx + logq. (4) q = xtjp 2 . 

(5) pe v = qe x . (6) q (p - cos x) = cos y. 

132. Staftd3t€4\L._Paftial differential equations analogous to Clair- 
aut's form. In Chap. VI. we showed that the complete primitive of 

y=px+f(p) 
was y — cx +f(c), a family of straight lines. 

Similarly the complete integral of the partial differential equation 
z=px+qy+f(p,q) 
is z=ax+by +f(a, b), a family of planes. 
For example, the complete integral of 
z=px +qy +p 2 +q 2 
is z — ax +by + a 2 + b 2 . 

Corresponding to the singular solution of Clairaut's form, giving 
the envelope of the family of straight lines, we shall find in the next 



PARTICULAR METHODS 155 

article a " singular integral " of the partial differential equation, 
giving the envelope of the family of planes. 

Examples for solution. 

(1) Prove that the complete integral of z=px + qy-2p-3q repre- 
sents all possible planes through the point (2, 3, 0). 

(2) Prove that the complete integral of z=px + qy + <\/(p 2 + q 2 + l) 
represents all planes at unit distance from the origin. 

(3) Prove that the complete integral of z=px + qy+pq/(pq-p-q) 
represents all planes such that the algebraic sum of the intercepts on 
the three co-ordinate axes is unity. 

133. Singular Integrals. In Chap. VI. we showed that if the 
family of curves represented by the complete primitive of an ordinary 
differential equation of the first order had an envelope, the equation 
of this envelope was a singular solution of the differential equation. 
A similar theorem is true eoncerning the family of surfaces repre- 
sented by the complete integral of a partial differential equation of 
the first order. If they have an envelope, its equation is called a 
" singular integral." To see that this is really an integral we have 
merely to notice that at any point of the envelope there is a surface 
of the family touching it. Therefore the normals to the envelope 
and this surface coincide, so the values of p and q at any point of 
the envelope are the same as that of some surface of the family, and 
therefore satisfy the same equation. 

We gave two methods of finding singular solutions, namely from 
the c-discriminant and from the ^-discriminant, and we showed that 
these methods gave also node-loci, cusp-loci, and tac-loci, whose 
equations did not satisfy the differential equations. The geometrical 
reasoning of Chap. VI. can be extended to surfaces, but the dis- 
cussion of the extraneous loci which do not furnish singular integrals 
is more complicated.* As far as the envelope is concerned, the 
student who has understood Chap. VI. will have no difficulty in 
understanding that this surface is included among those found by 
eliminating a and b from the complete integral and the two derived 
equations f(x,y,z,a,b)=0, 

da U ' 

db 

* See a paper by M. J. M. Hill, Phil. Trans. (A), 1802. 



156 DIFFERENTIAL EQUATIONS 

or by eliminating p and q from the differential equation and the 
two derived equations 

F(x, y, z, p, q) =0, 

dF =0 

dp u ' 

dF 

■ Bq 

In any actual example one should test whether what is apparently 
a singular integral really satisfies the differential equation. . 

Ex. (i). The complete integral of the equation of Art. 132 was 

z = ax + by + a 2 + b 2 . 
Differentiating with respect to a, = x + 2a. 
Similarly 0= y +26. 

Eliminating a and b, 4z = - (x 2 + y 2 ). 

It is easily verified that this satisfies the differential equation 
z=px + qtj + p 2 + q 2 
and represents a paraboloid of revolution, the envelope of all the planes 
represented by the complete integral. 

Ex. (ii). The complete integral of the equation of Art. 130 was 

9{x + ay + b) 2 = {z 2 + a 2 f (1) 

Differentiating with respect to a, 

18y{x + ay + b) = 6a{z 2 + a 2 ) 2 (2) 

Similarly 18(x + ay + b)=0 (3) 

Hence from (2), a = (4) 

Substituting from (3) and (4) in (1), z=0. 

But z = gives p = q = 0, and these values do not satisfy the differ- 
ential equation z 2 (p 2 z 2 + q 2 ) = l. 
Hence 2 = is not a singular integral. 

Ex. (iii). Consider the equation p 2 = zq. 

Differentiating with respect to p, 2p=0. 
Similarly = z. 

Eliminating p and q from these three equations, we get 

2 = 0. 

This satisfies the differential equation, so it really is a singular 
integral. 

But it is derivable by putting 6 = in 

z = be" x+ah J, 

which is easily found to be a complete primitive. 

So z-—0 is both a singular integral and a particular case of the 
complete integral. 



PARTICULAR METHODS 157 

Examples for solution. 

Find the singular integrals of the following : 

(1) z=px + qy + logpq. (2) z=px + qy + p 2 + pq + q 2 . 

(3) z=>px + qy + %p 2 q 2 . (4) z=px + qy+p/q. 

(5) iz = pq. (6) z 2 = l+p 2 + q 2 . (7) p 3 + ? 3 = 27z. 

(8) Show that no equation belonging to Standard I. or III. has a 
singular integral. [The usual process leads to the equation = 1.] 

(9) Show that z=0 is both a singular integral and a particular case 
of a complete integral of q 2 = z 2 p 2 (l -p 2 ). 

134. General Integrals. We have seen, in Ex. (i) of the last 
article, that all the planes represented by the complete integral 

z=ax+by + a 2 +b 2 (1 ) 

touch the paraboloid of revolution represented by the singular 
integral ± z = _( x z+ y 2y ■ (2) 

Now consider, not all these planes, but merely those perpendicular 
to the plane y =0. These are found by putting 6=0 in (1), giving 

z—ax + a 2 , 
of which the envelope is the parabolic cylinder 

4z = - x 2 (3) 

Take another set, those which pass through the point (0, 0, 1). 

From(l), l=a 2 +6 2 , 

so (1) becomes z=ax±y\/(l -a 2 ) +1, 

of which the envelope is easily found to be the right circular cone 

(z-l) 2 = x 2 +y 2 (4) 

In general, we may put b=f(a), where/ is any function of a, 
giving z=ax+yf(a)+a? + {f(a)} 2 (5) 

The envelope of (5) is found by eliminating a between it and 

the equation found by differentiating it partially with respect to a. 

i.e. 0=x+yf'{a)+2a+2f{a)f(a) (6) 

If / is left as a perfectly arbitrary function, the eliminant is 
called the " general integral " of the original differential equation. 
Fjquations (3) and (4) are particular integrals derived from the 
general integral. 

AVe may define the general integral of a partial differential 
equation of the first order as the equation representing the aggregate 
of the envelopes of every possible singly-infinite set of surfaces that 



158 DIFFERENTIAL EQUATIONS 

can be chosen out of the doubly-infinite set represented by the 
complete integral. These sets are denned by putting b -f (a) is 
the complete integral. 

It is usually impossible to actually perform the elimination of 
a between the two equations giving the envelope, on account of the 
arbitrary function /and its differential coefficient. The geometrical 
interest lies chiefly in particular cases formed by taking / as some 
definite (and preferably simple) function of a. 

135. Characteristics. The curve of intersection of two con- 
secutive surfaces belonging to any singly-infinite set chosen from 
those represented by the complete integral is called a characteristic. 

Now such a curve is found from the equation of the family of 
surfaces by the same two equations that give the envelope. For 
instance, equations (5) and (6) of the last article, for any definite 
numerical values of a,f(a), and /'(a), define a straight line (as the 
intersection of two planes), and this straight line is a characteristic. 
The characteristics in this example consist of the triply-infinite set 
of straight lines that touch the paraboloid of revolution (2). 

The parabolic cylinder (3) is generated by one singly-infinite set 
of characteristics, namely those perpendicular to the plane y=0, 
while the cone (4) is generated by another set, namely those that 
pass through the fixed point (0, 0, 1). Thus we see that the general 
integral represents the aggregate of all such surfaces generated by the 
characteristics. 

If a singular integral exists, it must be touched by all the char- 
acteristics, and therefore by the surfaces generated by particular 
sets of them represented by the general integral. It is easily verified 
that the parabolic cylinder and right circular cone of the last article 
touch the paraboloid of revolution. 

136. Peculiarities of the linear equation. To discuss the linear 

equation Pp+Qg^R (1) 

on these lines, suppose that u = const. 

and v = const, 

are two independent integrals of the subsidiary equations.* 
Then it is easily verified that an integral of (1) is 

■u +av+b=0 (2) 

* >Since u and v are independent, at least one of them must contain z. Let 
this one be u. We make this stipulation to prevent it +av + b being a function of 
x and y alone, in which case it +av + b=0 would make terms in (1) indeterminate, 
instead of definitely satisfying it in the ordinary way. 



PARTICULAR METHODS 159 

This may be taken as the complete integral. The general 
integral is found from 

\u+av+f(a)=0, (3) 

«+/»=0 (4) 

From (4), a is a function of v alone, 
say a = F(v). 

Substituting in (3), u =a function of v, 

say u = \}s(v), 

which is equivalent to the general integral </>(«, v)=0 found at the 
beginning of the chapter. 

The linear equation is exceptional in that its complete integral 
(2) is a particular case of the general integral. Another peculiarity 
is that the characteristics, which are here the curves represented by 
the subsidiary equations, are only doubly-infinite in number instead 
of triply-infinite. Only one passes through a given point (in general), 
whereas in the non-linear case, exemplified in the last article, an 
infinite number may do so, forming a surface. 

Examples for solution. 

(1) Find the surface generated by characteristics of 

z = px + qy + p 2 + pq + q 2 

that are parallel to the axis of x. Verify that it really satisfies the 
differential equation and touches the surface represented by the singular 
integral. 

(2) Prove that z 2 = 4xy is an integral of 

z=px + qy + log pq 

representing the envelope of planes included in the complete integral 
and passing through the origin. 

(3) Prove that the characteristics of q = 3p 2 that pass through the 
point (-1, 0, 0) generate the cone (x + l) 2 + 12yz = 0. 

(4) What is the nature of the integral (y + 1 ) 2 + \xz = 0of the equation 

'. 4 z = px + qy + p/q ? 

(5) Show that either of the equations 

z = {x + y) 2 + ax + by, 

. mx* + ny* 

z = (x + y) 2 + — 

J x + y 

may be taken as the complete integral of a certain differential equation, 
and that the other may be deduced from it as a particular case of the 
general integral, [London.] 



160 DIFFERENTIAL EQUATIONS 

(6) Show that z = (x + a) 2 e by is a complete integral of the differential 

equation p 2 = ±ze qylz . 

( xv \ 2 ~ v 
Show that y 2 z = i[~-) is part of the general integral of the 

same equation, and deduce it from the above given complete integral. 

[London.] 

MISCELLANEOUS EXAMPLES ON CHAPTER XII. 

(1) z=px + qy-p 2 q. (2) 0=px + qy-(px + z) 2 q. 

(3) z {z 2 + xy) (px - qy) = x i . (4) p*-q*=3x- 2y. 

(5) p 1 2 + 2x 2 p 2 + x 3 2 p 3 = 0. (6) x s p L + xtf> 2 + Xjp 3 =0. 

(7) p 3 + q*-3pqz = 0. (8) pS + p 2 2 +p 3 2 = ±z. 

(9) p x +Pi + p z = iz. (10) t> 2 + 6^ + 2? + 4=0. 

(11) z 2 p 2 y + 6zpxy + 2zqx 2 + 4x 2 y = 0. (12) zpy 2 = x(y 2 + z 2 q 2 ). 
(13) p 2 z 2 + q 2 = p 2 q. (14) (z-px-qy)x z y 2 = q 2 zx z -3pH 2 y 2 . 

(15) Find the particular case of the general integral of p + q=pq 
that represents the envelope of planes included in the complete integral 
and passing through the point (1, 1, 1). 

(16) Prove that if the equation P dx + Q dy + R dz = is integrable, it 
represents a family of surfaces orthogonal to the family represented by 

Pp+Qq = R. 
Hence find the family orthogonal to 

<p{z(x + y) 2 , x 2 -y 2 } = 0. 

(17) Find the surfaces whose tangent planes all pass through the 
origin. 

(18) Find the surfaces whose normals all intersect the circle 

x 2 + y 2 = l, 2 = 0. 

(19) Find the surfaces whose tangent planes form with the co- 
ordinate planes a tetrahedron of constant volume. 

(20) Prove that there is no non-plane surface such that every 
tangent plane cuts off intercepts from the axes whose algebraic sum is 
zero. 

(21) Show that if two surfaces are polar reciprocals with respect to 
the quadric x 2 + y 2 = 2z, and (x, y, z), (X, Y, Z) are two corresponding 
points (one on each surface) such that the tangent plane at either point 
is the polar plane of the other, then 

X = p; Y = q; Z=px + qy-z; x = P; y = Q. 
Hence show that if one surface satisfies 

./>, y, *, p, ?)=o, 

the other satisfies / (P, Q, PX +QY-Z, X. Y) = 0. 

(These equations are said to be derived from each other by the 
Principle of Duality.) 



MISCELLANEOUS EXAMPLES 161 

(22) Show that the equation dual to 

z=px + qy+pq 
ig 0=Z + XY, 

7)7 

giving x = P = 0x=-Y, y = Q=-X, 

z = PX+QY-Z=-XY. 

Hence derive (as an integral of the first equation) z= -xy. 



* CHAPTER XIII 

PARTIAL DIFFERENTIAL EQUATIONS OF THE FIRST 
ORDER. GENERAL METHODS 

137. We shall now explain Charpit's method of dealing with 
equations with two independent variables and Jacobi's method for 
equations with any number of independent variables. Jacobi's 
method leads naturally to the discussion of simultaneous partial 
differential equations. 

The methods of this chapter are considerably more complicated 
than those of the last. We shall therefore present them in their 
simplest form, and pass lightly over several points which might be 
considerably elaborated. 

138. Charpit's f method. In Art. 131 we solved the equation 

p-3x\ = q 2 -y (1) 

by using an additional differential equation 

p-Sx 2 =a, (2) 

solving for p and q in terms of x and y, and substituting in 

dz=pdx+qdy, (3) 

which then becomes integrable, considered as an ordinary differential 
equation in the three variables x, y, z. 

We shall now apply a somewhat similar method to the general 
partial differential equation of the first order with two independent 

variables F (x, y, z, p, q)=0 (4) 

We must find another equation, say 

f(x,y,z,p,q)=0, (5) 

* To be omitted on a first reading. 

f This method was partly due to Lagrange, but was perfected by Charpit.* 
Charpit's memoir was presented to the Paris Academy of Sciences in 1784, but 
the author died soon afterwards and the memoir was never printed. 

162 



GENERAL METHODS 163 

such that p and q can be found from (4) and (5) as functions of 
x, y, z which make (3) integrable. 

The necessary and sufficient condition that (3) should be in- 
tegrable is that 

^(dQ dR\ n (dR dP\ w /dP BQ\ A ,., „ . 

where p=^ Q=q } R= -l } 

r dz * dz dy dx ' v ' 

By differentiating (4) partially with respect to x, keeping y and 
z constant, but regarding p and q as denoting the functions of x, 
y, z obtained by solving (4) and (5), we get 

<W + dFdp + dFdq =0 
dx dp dx dqdx 

Similarly §? + £9e + ffj*_ (8) 

J dx dpdx dqdx 

-ci /»% J /ox rfy ^Fdf dFdf 

From (7) and (8), J £ = -J--^ £, (9) 

U I ' * A t dFd f dFd f 

where J stands for =- #- - =- ^- . 

dp dq dq dp 

J%- f |-|| (10) 

J df = JFdf + dFdf 

dy dy dq dqdy' 
dp_ dFdf dFdf 
J dz-~dzdq*TqJ* (12) 

Substituting in (6) and dividing out * <7, we get 

(dF^f_dFdf\ fdFdf_dFdf\ 
*\dz dp dpdzJ ^Vdz dq dq dz ) 

dF^f_dF^f + dF^f_dFBf_ 
dy dq dq dy dx dp dp dx ' 

dpdx dqdy \ dp ^ dq) dz 

(dF dF\df (dF dF\df A , lox 
S + {Tx + P-dz)dp + [dy-^dz)drr - "'<*) 

* J cannot vanish identically, for this would imply that F and /, regarded as 
functions of p and q, were not independent. This is contrary to our hypothesis 
that equations (4) and (5) can be solved for p and q. 



164 DIFFERENTIAL EQUATIONS 

This is a linear equation of the form considered in Art. 126, 
with x, y, z, p, q as independent variables and / as the dependent 
variable. 

The corresponding subsidiary equations are 

dx dy dz dp dq _df ,,.* 

"JF~'JF~ _ dF_ d_F~d_F dlTdF dF~6' { ' 
dp dq dp ^ dq dx " dz dy ^ dz 

If any integral of these equations can be found involving p or 
q or both, the integral may be taken as the additional differential 
equation (5), which in conjunction with (4) will give values of p 
and q to make (3) integrable. This will give a complete integral of 
(4), from which general and singular integrals can be deduced in 
the usual way. 

139. As an example of the use of this method, consider the 
equation 2xz-px 2 -2qxy +pq=0 (1) 

Taking the left-hand side of this equation as F, and substituting 
in the simultaneous equations (14) of the last article, we get 

dx dy dz dp dq _df 

x 2 - q 2xy -p px 2 + 2xyq - 2pq 2z -2qy ' 

of which an integral is q = a (2) 

2x{z-ay) 



From (1) and (2), p = : 



ar -a 



1 , 7 2x(z-ay)dx 7 

Hence dz=pdx+qdy= — —^ — — — +ady, 

0dz -ady _ 2x dx 

1.€. — s » 

z - ay x z - a 
i.e. z = ay + b (x 2 - a) . 
This is the complete integral. It is easy "to deduce the Singular 
Integral z = x 2 y. 

The form of the complete integral shows that (1) could have 
been reduced to z = PX + qy - Pq, 

which is a particular case of a standard form, by the transformation 

X 2= X . p= dz_^dz 
' dX 2x dx' 

Equations that can be solved by Charpit's method are often 
solved more easily by some such transformation. 



GENERAL METHODS 165 

Examples for solution. 

Apply Charpit's method to find complete integrals of the following : 
(1) 2z+p 2 + qy + 2y 2 =0. (2) yzp 2 = q. 

(3) pxy+pq + qy = yz. (4) 2x(z 2 q 2 + l)—pz. 

(5) 3 = 3^ 2 . (Cf. Art. 129.) (6) z 2 (p 2 z 2 + q 2 ) = l. (Cf. Art. 130.) 

(7) p-3x 2 = q 2 -y. (Cf. Art. 131.) 

(8) z=px + qy+p 2 + q 2 . (Cf. Art. 132.) 

(9) Solve Ex. 2 by putting y 2 = Y, z 2 =Z. 

(10) Solve Ex. 4 by a suitable transformation of the variables. 

140. Three or more independent variables. Jacobi's* method. 

Consider the equation 

F(x 1} x 2 , x 3 , ft, ft, ft) =0, (1) 

where the dependent variable z does not occur except by its partial 
differential coefficients ft, ft, p 3 with respect to the three independent 
variables x v x 2 , x 3 . The fundamental idea of Jacobi's method is 
very similar to that of Charpit's. 

We try to find two additional equations 

F x {x x , x 2 , Z3, ft, ft, ft)=a l5 (2) 

F 2 (x v x 2 , x 3 , p x ,p % ,Pz)=a 2 (3) 

(where a x and a 2 are arbitrary constants), such that ft, p 2 , p 3 can 
be found from (1), (2), (3) as functions of x 1} x 2 , x 3 that make 

dz =p 1 dx 1 +p 2 dx 2 +p 3 dx 3 (4) 

integrable, for which the conditions are 

d?2 = d 2 z ^dft. 3ft = 3ft. dft = 3ft ^ 

dx 1 dx x dx 2 dx 2 dx x dx 3 dx 2 dx 3 " 

Now, by differentiating (1) partially with respect to x v keeping 
x 2 and x 3 constant, but regarding ft, ft, ft as denoting the functions 
of x v x 2 , x 3 obtained by solving (1), (2), (3), we get 

01 + dF d h + ?I fe + ^ ^3 =0 (6) 

dx x 3ft dx x 3ft dx x 3ft 3^ 

Similarl d Il + d Jj d JPl + d L^ + d Il d h ^o (7) 

^ dx x 3ft dx x 3ft 3^ 3ft dxj 

* Carl Gustav Jacob Jacobi of Potsdam (1804-1851) may bo considered as one 
of the creators of the Theory of Elliptic Functions. The " Jacobian " or " Func- 
tional Determinant " reminds us of the large part he played in bringing deter- 
minants into general use. 



166 DIFFERENTIAL EQUATIONS 

From (6) and (7), 

d(F, F x ) | d(F, F x \ dp 2 | d(F, F x ) dp z _ 8) 

d(*i>Pi) d(P2,Pi) fax 3{VvVi) dx i ' "" 

where ^7— - — H denotes the " Jacobian" =— 5-^-s— =-*. 
dO&i> ^1) ^1 ?Pi "Pi ™h. 

Similarly 

d(F,F x ) ^ (FtFJ dft ^(FtFJ dfr^ (g) 

d {x 2 , p 2 ) d (p v p 2 ) dx 2 d (p 3 , p 2 ) dx 2 '"' 

and d(F, F x ) ^(F.FJ dp, | d(F, F x ) dp, =Q 

^ (a*> ? 3 ) 3 (p 1} #,) ax 3 9 (p 2 , p 3 ) dx 3 
Add equations (8), (9), (10). 
Two terms are 

d(F, F x ) dp 2 | d(F, F x ) d Pl _ dH mF,F x ) } d(F, F t ) \ =0 
d (P& Pi) dx i d(Pv Pt) dx 2 dx x dx 2 Xd{p 2 , p x ) d{p v p 2 )j 
Similarly two other pairs of terms vanish, leaving 

3(2?,^)^,^) | 3(J , ,J , i) H) (n) 

d(x 1 ,p 1 ) d(x 2 ,p 2 ) d(x 3 ,p 3 ) ' 

i e dFdJ\JFdJ\ + ^BF 1 _dFdJ\ + dFdJ\_dFdF 1=0 
dx x dp x dp x dx x dx 2 dp 2 dp 2 dx 2 dx 3 dp 3 dp 3 dx 3 

This equation is generally written as (F, F x ) =0. 

Similarly (F, F 2 )=0 and (F x , F 2 ) =0. 

But these are linear equations of the form of Art. 126. Hence 
we have the following rule : 

Try to find two independent integrals, F x = a x and F 2 = a 2 , of the 
subsidiary equations 

dx x dp x dx 2 dp 2 dx 3 dp 3 

"JIT tlT ~JL = W. = 'j F== W 
dp x dXf, dp 2 dx 2 dp 3 dx 3 

If these satisfy the condition 

IF SM-V^ 9 ^ dF i dF *\-0 
{*»**)=Zi\ fa r dp~ r ~dp~ r dx~ r )-^ 

and if the p's can be found as functions of the x's from 

F = F x -a x = F 2 -a 2 =0, 
integrate the equation* formed by substituting these functions in 

dz =p 1 dx 1 +p 2 dx 2 +p 3 dx 3 . 

* For a proof that this equation will always be integrable, see Appendix C. 



GENERAL METHODS 167 

141. Examples on Jacobi's method. 

Ex. (i). 2p 1 x 1 x 3 + Sp 2 x 3 2 +p 2 2 p 3 = (1) 

The subsidiary equations are 

dx 1 dp 1 _ dx 2 dp 2 dx 3 dp 3 
-2xjX 3 2p^ 3 ~ -3x 3 2 -2p 2 p 3 ~ ~ -p 2 2 ~2p l x 1 + 6p 2 x 3 
of which integrals are F 1 =p 1 x i =a v (2) 

and F2-P2 = a z (3) 

Now with these values (F v F 2 ) is obviously zero, so (2) and (3) can 
be taken as the two additional equations required. 

p 1 = a 1 x 1 -\ p 2 = a 2 , p 3 = -a 2 - 2 (2a 1 x 3 + 3a 2 x 3 2 ). 
Hence dz = a l x l ~ 1 dx l + a 2 dx 2 - a 2 ~ 2 {2a 1 x 3 + 3a 2 x 3 2 ) dx 3 
or z = a 1 log x 1 + a 2 x 2 - a 2 ~ 2 (a x x 3 2 + a 2 x 3 3 ) + a 3 , 

the complete integral. 

Ex. (ii). {x 2 + x 3 ){p 2 +p 3 ) 2 + z Vl =0 (4) 

This equation is not of the form considered in Art. 140, as it involves 
2. But put 

dz dx x du I du „ ._ 

•-** ^wr^r-dx-Jdx-r J 4,say ' 

where u=0 is an integral of (4). 

Similarly p 2 = - PJP t ; p z = - P 3 /P 4 . 

(4) becomes (x 2 + x 3 ){P 2 + P 3 ) 2 -x i P 1 P 4 = 0, (5) 

an equation in four independent variables, not involving the dependent 
variable u. 

The subsidiary equations are 

dx 1 dP 1 dx 2 _ dP 2 _ dx 3 

V> 4 = ~0~ = -2(x 2 + x 3 )(P 2 + P 3 ) = (P 2 + P 3 ) 2== -2(x 2 + x 3 )(P 2 + P 3 ) 
dP 3 dx 4 dP i 

of which integrals are F 1 =P 1 = a v (6) 

F 2 =P 2 -P 3 = a 2 (7) 

F 3 =x 4 Pi = a 3 (8) 

We have to make sure that (F r , F s ) = 0, where r and s are any two 
of the indices 1, 2, 3. This is easily seen to be true. 
Solving (5), (6), (7), (8), we get 

P 1 = a 1 ; P i = a 3 x i ~ 1 ; 2P 2 = a 2 ±\/{a 1 a 3 /(x 2 + x 3 )} ; P 3 = P 2 -«2> 

so du = a l dx l + a^f 1 dx 4 + \a 2 (dx 2 - dx 3 ) 

± W{ a l a J( X i + X 3)} (J X 2 + ^3), 

i.e u = a x x x + a 3 log z 4 + \a 2 {x 2 - x 3 ) ± y/{a 1 a 3 (x l + ^3)} + °4- 



168 DIFFERENTIAL EQUATIONS 

So m=0 gives, replacing 

%4 by z , a il a 3 by A v W a 3 by A 2 , aja 3 by A 3 , 

log Z + A X X X + A 2 (X 2 - X 3 ) ± ^{A x {x 2 + ^s)} + ^3 = 0, 

the complete integral of (4). 

Examples for solution. 

Apply Jacobi's method to find complete integrals of the following : 

(1) Pi* + p 2 2 + p 3 = l. (2) ^Vl'aV+ftV-}'^ ' 

(3) p x x x +p 2 x 2 =p 3 2 . (4) p x p 2 p 3 +p 4 s x J x 2 x 3 x^=0. 

(5) p x p$> 3 = z z x x x 2 x 3 . (6) p 3 x 3 (p x +p 2 )+x x + x 2 =0. 

(7) p x 2 +p 2 p 3 -z{p 2 + p 3 )=0. 

(8) (j? 1 + r» 1 ) 2 + (^ 2 + x 2 ) 2 + (^3 + a;3) 2 = 3(a; 1 + a; 2 + a;3). 

142. Simultaneous partial differential equations. The following 
examples illustrate some typical cases : 

Ex. (i). F=p x 2 + p 2 p 3 x 2 x 3 2 = 0, fl) 

F l =p 1 +p 2 x 2 =0 (2) 

Here 

Thus the problem may be considered as the solution of the equation 
(1), with part of the work (the finding of F x ) already done. 
The next step is to find F 2 such that 

(F,F 2 )=0 = (F x ,F 2 ). 

The subsidiary equations derived by Jacobi's process from F are 

dx x dp x dx 2 dp 2 dx 3 dp 3 ■ 

-%Pi~ -P3^2 X 3 2 ~P2V3 X 3 2 ~ -p 2 X 2 X 3 2 ~2p 2 ]) 3 X 2 X 3 

An integral is Pi~ a (3) 

We may take F 2 as p x , since this satisfies (F, F 2 )=0 = (F v F 2 ). 
Solving (1), (2), (3) and substituting in dz=p 1 dx 1 +p 2 dx 2 + p 3 dx 3 , 

dz = a dx x - gkk 2 -1 dx 2 + ax 3 ~ 2 dx 3 , 
so z = a(x x -\ogx 2 -x 3 ~ 1 ) + b. 

Ex. (ii). F==p x x x + P2 x 2 -p 3 *=0, (4) 

F x = Pi p 2 +p 3 -l=0 (5) 

Here (F, F x )=p x + p 2 -l) = p x -p 2 . 

This must vanish if the expression for dz is to be integrable. 

Hence we have the additional equation 

Pi-P 2 =0 (6) 

Solving (4), (5), (6) and substituting, 

dz = — + dx 3 , 

z = log (cc j + x 2 ) +x 3 + a. 






GENERAL METHODS 169 

In examples of this type we do not have to use the subsidiary 
equations. The result has only one arbitrary constant, whereas in 
Ex. (i) we got two. 

Ex. (iii). F=x x 2 + x 2 2 +p 3 =0, (7) 

Jispi+i> i +*, , -0 (8) 

Here (F, FJ = 2x x + 2x 2 - 2x 3 . 

As x v x 2 , % 3 are independent variables, this cannot be always zero. 
Hence we cannot find an integrable expression for dz from these 
equations, which have no common integral. 

Ex. (iv). F=p 1 +p 2 +p 3 2 -3x 1 -3x 2 -ix 3 2 = 0, (9) 

F l =x 1 p 1 -x 2 j) 2 -2x 1 2 + 2x 2 2 = 0, (10) 

F 2 =p 3 ~2x 3 =0. (11) 

Solving (9), (10), (11) and substituting in the expression for dz, 

dz = (2x t + x 2 ) dx x + (x x + 2x 2 ) dx 2 + 2x 3 dx 3 , 

so z = x l 2 + x l x 2 + x 2 2 + x 3 2 + a. 

This time there is no need to work out (F, F x ), (F, F 2 ), {F x , F 2 ). 

Ex. (v). F= Pi +p 2 -l-x 2 = 0, (12) 

F 1 =p l +p 3 -x 1 -x 2 = 0, (13) 

F 2 =p 2 +p s -l-x 1 =0 (14) 

These give dz = x z dx x + dx 2 + x x dx 3 . 

As this cannot be integrated, the simultaneous equations have no 
common integral. 

Ex. (vi). F=x 1 p x -x 2 p 2 + p 3 -p i =0, (15) 

F 1 =sp l + p 2 -x 1 -x 2 =:0 .....(16) 

Here (F, F l )=p 1 - x 1 {-l)-p 2 + x 2 {-l)=p l -p 2 + x 1 -x 2 . 

As in Ex. (ii), this gives us a new equation 

F 2 =p 1 -p 2 + x 1 -x 2 =-0 (17) 

Now (F, F 2 )=p 1 -x 1 -p 2 (-l)+x 2 (-l) = F 1 =Q, 

and (F x , ig = (-l)-l+(-l)(-l)-(-l)=0, 

so we cannot get any more equations by this method. 
The subsidiary equations derived from F are 

dx x dp x dx 2 _ dp 2 _dx 3 _dp 3 _dx i _dj) 4 
-x x ~ p x ~ x 2 -p 2 -1 1 6 

A suitable integral is F 3 = p 3 = a, (18) 

for this satisfies (F, F 3 ) = (F V F 3 ) = (F 2 , F 3 )=0. 

We have now four equations (15), (16), (17), (18). These give 

Pi = z 2 ; P 2 = a; i; Vz = a "> V^ a '> 
so z = x x x 2 + a (x 3 + :c d ) + b. 



170 DIFFERENTIAL EQUATIONS 

But in this example we can obtain a more general integral. The 
two given equations (15) and (16) and the derived one (17) are 
equivalent to the simpler set : 

Vi= x z> ' < 19 ) 

? 2 = z,> (20) 

?3-?4=0 (21) 

From (19) and (20), z = x l x 2 + a,ny function of x 3 and x 4 . 

(21) is a linear equation of Lagrange's type, of which the general 
integral is <}>(z, x 3 + %i)=0, 

i.e. z is any function of {x 3 + x 4 ), and may of course also involve a;, 
and x 2 . 

Hence a general integral of all three equations, or of the two given 
equations, is z = XyX ^ + ^ (^ + x ^ 

involving an arbitrary function. The complete integral obtained by 
the other method is included as a particular case. The general integral 
could have been obtained from the complete, as in Art. 134. 

Examples for solution. 

Obtain common complete integrals (if possible) of the following 
simultaneous equations : 

(1) Pl 2+p 2 2-8( Xl + X2 )2=0, 

(p 1 -p 2 )(x 1 -x 2 )+p 3 x 3 -l=0. 

(2) aj 1 a p a p8= 3 ®2 a PaPi = a; 3 2 7'iy2 = 1 - 

(3) PiP^Pa- 8x^X3=0, (4) 2x 3 p 1 p 3 -x i p i = 0, ' 
p 2 + p 3 -2x 2 -2x 3 =0. 2p t -p 2 =0. 

(5) p x x 3 2 + p 3 =0, (6) p 2 2 +p 3 3 + x 1 + 2x 2 + 3x 3 =0, 

p& a *+p&t* = p 1 +p 4 2 x 4 -l=0. 

(7) 2p 1 +p 2 +p 3 + 2p i =0, 
PlP3-P2?4=0. 

(8) Find the general integral of Ex. (5). 

(9) Find the general integral of Ex. (7). 

MISCELLANEOUS EXAMPLES ON CHAPTER XIII. 

(1) 2x y x 3 zp l p 3 + x 2 p 2 = 0. (2) x 2 p 3 + x 1 p i =p 1 p 3 -p 2 p 4 + x 4 2 =0. 

(3) 9x 1 x i p 1 (p 2 + p 3 )-ip 4 2 =0, (4) 9x l zp 1 (p 2 + p 3 )-4:=0 ) 

p 1 x 1 +p 2 -p 3 =0. PiXi+P2-Pa = Q- 

(0) x 1 p 2 p 3 = x 2 p 3 p J —x 3 p^p 2 = z x±x 2 x 3 . 

(6) p 1 z 2 -x 1 2 =p 2 z 2 -x 2 2 =p 3 z 2 -x 3 2 = 0. 

(7) Find a singular integral of z — p l x 1 + p 2 x 2 + p 3 x 3 + p t 2 + p 2 2 + p 3 2 , 
representing the envelope of all the hyper-surfaces (in this case hyper- 
planes) included in the complete integral. 

(8) Show that no equation of the form F(x v x 2 , x 3 , p v p 2 , p 3 )=0 
has a singular integral. 



Miscellaneous examples 171 

(9) Show that if z is absent from the equation F{x, y, z, p, q)-0, 
Charpit's method coincides with Jacobi's. 

(10) Show that if a system of partial differential equations is linear 
and homogeneous in the p's and has a common integral 

z = a 1 u l + a 2 u 2 + ... , 
where the u'b are functions of the sc's, then a more general integral is 

z = <t>(u v w a , ...)• 
Find a general integral of the simultaneous equations 
x 1 p 1 -x 2 p 2 + x 2 p 3 =0, 
x i p 3 -x^p i + x 5 p b = 0. 

(11) If p l and p 2 are functions of the independent variables x v x 2 
satisfying the simultaneous equations 

F(x v x 2 , p v p 2 )=0 = F 1 (x 1 , x 2 , p v p 2 ), 

proved (r.MkW^r- 

Hence 3how that if the simultaneous equations, taken as partial 
differential equations, have a common integral, (F, F x ) =0 is a necessary 
but not a sufficient condition. 

Examine the following pairs of simultaneous equations : 
(i) 2^^ + 2^-2 = 0, 

2^(^+2^-1=0. 

d(F F-.) 
CHere ~ — = identically, and the equations cannot be solved 

d(Vvlh) 
\ for p x and p 2 .] 

\ (ii) F^ Pl -p 2 * = 0, 

\ F 1 =p 1 + 2p z x x +x 1 i =0. 

d(F F.) 
THere (F, F-.) and ^ — '- — ^ both come to functions which vanish 
d(Pv Vi) 
when the p's are replaced by their values in terms of x x and x 2 There 
is no common integral.] 

(iii) F= Pl -p 2 2 + x 2 =0, 

F x ^p x + 2^2^! + x x 2 + x 2 =0. 

d(F, F,) 
[These have a common integral, although _. - comes to a 

d(Pi> P-i) 
function that vanishes when the p's are replaced by their values.] 



CHAPTER XIV 

PARTIAL DIFFERENTIAL EQUATIONS OF THE SECOND 
AND HIGHER ORDERS 

143. We shall first give some simple examples that can be 
integrated by inspection. After this we shall deal with linear 
partial differential equations with constant coefficients ; these are 
treated by methods similar to those used for ordinary linear equations 
with constant coefficients. The rest of the chapter will be devoted 
to the more difficult subject of Monge's* methods. It is hoped that 
the treatment will be full enough to enable the student to solve 
examples and to make him believe in the correctness of the method, 
but a discussion of the theory will not be attempted.! 

Several examples will deal with the determination of the arbitrary 
functions involved in the solutions by geometrical conditions. J 

The miscellaneous examples at the end of the chapter contain 
several important differential equations occurring in the theory of 
vibrations of strings, bars, membranes, etc. 

d 2 z d 2 z d 2 z 
The second partial differential coefficients ^~ v , - , ~~-% will 

be denoted by r, s, t respectively. 

144. Equations that can be integrated by inspection. 

Ex. (i). s = 2x + 2y. 

Integrating with resj)ect to x (keeping y constant), 

q = x 2 + 2xy + e/)(y). 
Similarly, integrating with respect to y, 

z = x 2 y + xy 2 + I (y) dy + f(.r), 

say z = x 2 y + xy 2 +f (x) + F(y). 

* Gaspard Monge, of Bcaune (1746-1818), Professor at Paris, created Descriptive 
Geometry. He applied differential equations to questions in solid geometry. 

t The student who desires this should consult Goursat, Sur V integration des 
equations mix derivees partielles du second ordre. 

X Frost's Solid Geometry, Chap. XXV., may be read with advantage. 

172 



SECOND AND HIGHER ORDERS 173 

Ex. (ii). Find a surface passing through the parabolas 
z = 0, y 2 = iax and 2 = 1, y z = -iax, 
and satisf ying xr + 2p=0. 

The differential equation is 

giving x 2 p=f(y), 

z=-\f(y) + F{y). 

The functions / and F are to be determined from the geometrical 
conditions. 

Putting 2 = and x = y 2 /ia, 

Similarly l=~f(y) + F(y). 

Hence F(y)J-, /(</)=£ 



and 



Z ~2 8acc' 



i.e. 8a:r2: = 4ax - ?/ 2 , a conicoid. 

Examples for solution. 

(1) r = 6a;. (2) xys = l. 

(3) £ = sina^/. (4) xr + p = 9x 2 y 3 . 

(5) 2/s +^> = cos (x + y) -y sin (» + ?/). (G) t-xq = x 2 . 

(7) Find a surface satisfying s = 8xy and passing through the circle 

z=0 = x 2 + y 2 -l. 

(8) Find the most general conicoid satisfying xs + q = Ax + 2y + 2. 

(9) Find a surface of revolution that touches 2 = and satisfies 

r = 12x 2 + 4y 2 . 

(10) Find a surface satisfying t = 6x 3 y, containing the two lines 

y = = z, y = \=z. 

145. Homogeneous linear equations with constant coefficients. Tn 
Chap. III. we dealt at some length with the equation 

(D n +a 1 D n - 1 +a 2 D»- 2 + ... +a n )y=f(x), (1) 

where 7)=^-. 
ax 



174 DIFFERENTIAL EQUATIONS 






We shall now deal briefly with the corresponding equation in 
two independent variables, 

(Z>« + aJ) n -W + a 2 D n -*D' 2 + ...+ a n D' n ) z =f(x, y), (2) 

where Z)e= — , D'==- . 

ox oy 

The simplest case is (D - mD')z =0, 

i.e. p-mq=0, 

of which the solution is (p{z, y+mx) =0, 

i.e. z=F(y +mx). 
This suggests, what is easily verified, that the solution of (2) 
iif{x,y)=0is 

z = F 1 (y+ mp) + F 2 (y + m&) + ...+F n (y + m n x), 
where the %, m 2 , ... m n are the roots (supposed all different) of 
m n +a 1 m n - 1 +a 2 m n ~ 2 + ... +a n =0. 

3^z o z o z 

dx? dx 2 dy dxdy 2 ' 
i.e. (Z) 3 -3D 2 Z)'+2DZ)' 2 )2=0. 
The roots of m 3 -3m 2 + 2m=0 are 0, 1, 2. 
Hence z = F 1 (y) + F 2 (y + x) + F 3 (y + 2x). 

Examples for solution. 

(1) (IP-GDW + UDD' 2 -6D' 3 )z=0. 

d 2 z d 2 z 

(2) 2r + 5s + 2*=0. (3) g^-gp-0. 

(4) Find a surface satisfying r + s = and touching the elliptic 
paraboloid z = 4x 2 + y 2 along its section by the plane y = 2x + 1 . [N. B. — 
The values of p (and also of q) for the two surfaces must be equal for 
any point on y = 2x + l.] 

146. Case when the auxiliary equation has equal roots. Consider 
the equation (D-mD') 2 z=0 (1) 

Put (D-mD')z=u. 

(1) becomes (D -mD') w=0, 

giving u=F(y + mx) ; 

therefore (D -mD')z=F(y +mx), 

or p-mq = F(y + mx). 

The subsidiary equations are 

dx _ dy dz 

1 -m F(y+mx)' 



SECOND AND HIGHER ORDERS 175 

giving y +mx=a, 

and • dz-F(a)dx=0, 

i.e. z-xF (y + mx) = 6, 
so the general integral is 
<f>{z-xF(y+mx), y+mx}=0 or z=xF(y +mx) + F x (y +mx). 
Similarly we can prove that the integral of 
(D-mD') n z=0 
is z=x n ~ l F(y +mx) +x n ~ 2 F 1 (y +mx) + ... +F n _ 1 (y +mx). 

Examples for solution. 

(1) (4D 2 + 12DD'+9D'2)z=0. (2) 25r-40s + 16t =0. 

(3) (Z>»-4Z) a Z)'+4Z)Z)' a )2=0. 

(4) Find a surface passing through the two lines 2 = 2 = 0, 
2-1 =x-y=0, satisfying r-is + it =0. 

147. The Particular Integral. We now return to equation (2) of 
Art. 145, and write it for brevity as 

F(D,D')z=f(x,y). 

We can prove, following Chap. III. step by step, that the most 
general value of z is the sum of a particular integral and the 
complementary function (which is the value of z when the differ- 
ential equation has /(a?, y) replaced by zero). 

The particular integral may be written — ^- r .f(x, y), and 

we may treat the symbolic function of D and D' as we did that of 
D alone, factorising it, resolving into partial fractions, or expanding 
in an infinite series. 

1 1 / 87)'\- 2 

E ' 9 ' D* -6DD' +9^, (12^+36^) = F> (l --^-j (12tf + 36xy) 

= F 2 ( 1+ ^ +27 ^ 2 + -)' (12a;2 + 3G ^ ) 

=-^ 2 . (12a; 2 + S6xy) + ~ . 3Qx 

= x 4 + Qx?y + 9x 4 = lOz 4 + 6x 3 y, 
so the solution of (D 2 - WD' + 9D' 2 ) z = l2x 2 + S6xy 
is 2 = 10a 4 +6x 3 y + <p(y +3x) + x\Js(y +'3x). 

Examples for solution. 

(1) (D 2 -2DD' + D' 2 )z = l2xy. 

(2) {2D 2 -5DD' +2D' 2 )z = 2i(y-x). 



176 DIFFERENTIAL EQUATIONS 

(3) Find a real function V of x and y, reducing to zero when y=0 
and satisfying Q 2 y ^ 2 y 

_ + —„-4x<** + ,,*). 

148. Short methods. When / (x, y) is a function of ax + by, 
shorter methods may be used. 

Now D(p (ax + by) = a<p' (ax + by) ; D'<f> (ax + by) = b<p' (ax + by). 

Hence F (D, D') <j> (ax + by) = F (a, b) (f> {n) (ax + by), 
where <f> {n) is the n th derived function of <j>, n being the degree of 
F(D, D'). 

Conversely 

F (D, U) ^ ( aX + by ^ = Fja^b) $ ^ + ^ (A) 

provided F (a, 6)=/=0, e.g. 

1 .„ „ >_ -sin(2a:+3y) 

D*-4D*D' +4DD' 2 ° OS ^ x+6y) ~2 3 -4 . 2 2 . 3 +4 . 2 . 3 2 

= -32sin(2aj+3y), 

since (f> (2x + 3</) may be taken as - sin (2x + 3y) if 
<p'" (2x +3y) =cos (2x +3y). 
To deal with the case when F(a, b) = 0, we consider the equation 
(D - mB') 2sp- mq = x r \fr(y + mx), 
of which the solution is easily found to be 

x r+l 

so we may take 

1 x r+1 

D _ mD > • xr ^(y + mx ) =f+i ^(y+™x). 

Hence 

t^ ^jz-\lr(v +mx) = - rF , ■„,. , .x\}/(y +mx) = ... 

(D-mD) nYKJ ' (D-mD) n - 1 v XJ ' 

x n 

=~,i'(y+ mx )> ( B ) 

e ' 9 ' D 2 -2DD , +D' 2 tan (?/ + ^ = ^ tan ^ + ^' 
while 

D>-5DD' + W> Sin (4 * + y) = irm ■ D^D' sin {ix +y) 

= i) _ 4/)' • - 4 cos ( 4x + y"> by (A) 

= -\x cos (4a: + y) by (B). 



SECOND AND HIGHER ORDERS 177 

Examples for solution. 

(1) (D 2 -2DD' + D' 2 )z = e x + 2 v. 

(2) (D*-6DD' +9D' 2 )z = 6x + 2y. 

(3) (D^-4D 2 D' + ±DD' 2 )z=i sin (2x + y). 

d 2 V d 2 V 

(4) ar-.-ai-fc"/* (5) ^ + °^=l2(x + y) 

(6) 4r-4s + « = 161og(a; + 2?/). 

149. General method. To find a general method of getting a 
particular integral, consider 

(D -mD')z = p-mq=f(x, y). 
The subsidiary equations are 

dx _ dy _ dz 
1 ~ -m~f(x, yY 
of which one integral is y +mx=c. 

Using this integral to find another, 
dz =/ (x, c - mx) dx, 

z = I f(x, c-mx) dx + constant, 

where c is to be replaced by y - mx after integration. 

Hence we may take ^ jy .f(x, y) as I f(x, c -mx) dx, where 

c is replaced by y + mx after integration. 

Ex. (D-2D')(D + D')z = (y-l)e*. 

Here \f{x, c-2x)dx=\ (c-2x-l)e x dx = {c-2x + l)e x . 

Therefore -=r — —=-, . (y - 1) e x = (y + 1) e x , replacing c by y + 2x. 

Similarly jr — =-, . {y + l)e x is found from I (c + x + l)e x dx = (c + x)e* 

by replacing c by y — x, giving ye x , which is the particular integral 
required. 

Hence z=^ye x + (p{y + 2x)+\fs(y -x). 

Examples for solution. 

(1) (D 2 + 2DD' + D' 2 )z = 2cosy-XKh\y. 

(2) (D 2 -2DD' -WD' 2 )z = 12xy. (3) (r + s-Gt)z = y cos x. 

(4) ^-^- 2 ^ = ^ + ^-y«)8m^-co8^. 

(5) r-t = tan 3 x tan y - tan x tan 3 ?/. 

{ ' dx 2 dt 2 t 2 x 2 ' 

P.D.E. M 



178 DIFFERENTIAL EQUATIONS 

150. Non-homogeneous linear equations. The simplest case is 

(D-mD' -o)2=0, 

i.e. p - mq = az, 
giving <p (ze~ ax , y+mx) =0, 

or z=e ax \fs(y +mx). 

Similarly we can show that the integral of 

(D-mD' -a)(D-nD' -b)z=0 
is z=e ax f(y+mx)+e bx F(y+nx), 

while that of (D - mD' - afz = 

is z=e ax f(y+mx)+xe ax F(y+mx). 

But the equations where the symbolical operator cannot be 
resolved into factors linear in D and D' cannot be integrated in this 
manner. 

Consider for example (D 2 -D')z=0. 

As a trial solution put z = e 1uc+ky , giving 

{D 2 -D')z = (h 2 -k)e'" +ki >. 

So z=e h{x+hy) is a particular integral, and a more general one is 
Y l Ae h(xJrliy \ where the A and h in each term are perfectly arbitrary, 
and any number of terms may be taken. 

This form of integral is best suited to physical problems, as was 
explained at some length in Chap. IV. Of course the integral of 
any linear partial differential equation with constant coefficients 
may be expressed in this manner, but the shorter forms involving 
arbitrary functions are generally to be preferred. 

Examples for solution. 

(1) DD'(D-2D' -3)2 = 0. (2) r + 2s + t + 2p + 2q + l=0. 

(3) li* Tt- (4) ( D2 - I) ' 2+I) - I) ') z = - 

d 2 V d 2 V 
(5) (2D4-3D 2 Z)' + Z)' 2 )z=0. (6) ^T+^F=» 2 ^ 

(7) (Z)-2D'-l)(Z>-2Z>' 2 -l)z = 0. 

(8) Find a solution of Ex. (4) reducing to 1 when x= +00 and to 
?/ 2 when x = 0. 

151. Particular Integrals. The methods of obtaining particular 
integrals of non-homogeneous equations are very similar to those in 
Chap. III., so we shall merely give a few T examples. 

Ex. (i). (IP-3DD' + D + l)z = e 2 *+ s y, 

i r 2x+3y 

e 2x+3?/ = = _ l_ e 2x+Sy 



P i -3DD' + D + ] ' 2 3 -3.2.3+2 + l 



SECOND AND HIGHER ORDERS 179 

Hence z = - }e 2x+3 v + lAe hx+k ^, 

where h* - 3hk +h + 1 =0. 

Ex. (ii). (D + D'-l){D + 2D'-3)z = 4: + 3x + 6y. 

= i{l + D + D' + terms of higher degree} 

( D + 2D' . L . , J ) 

x 1 1 H 1- terms of higher degree J- 

. j. 4D + 5Z)' ... , , ) 

= -g- 1 1 H h terms of higher degree j-. 

Acting on 4 + 3x + 6y, this operator gives 

${4+3x + 6y+4 : + 10}=>6 + z + 2y. 
Hence z = 6 + x + 2y + e x f (y-x) + e 3x F(y - 2x). 

Ex. (Hi). (Z) 2 - DD f - 2D) z = sin (3x + 4y). 

D 2 -DD'-2D • Sin (3 * + 4y) = -32-(-3.4)-2D ' "^ f ^ + iy) 

= 3^2^* sin(3aJ + 4 ^ 
_ 3 + 2J 3 sin (3s + 4y) + 6 cos (8s + 4y) 

-9T4^- sin(,3a; + 4 ^-~ 9 -4(-3 2 ) 

= T V sin (3x + 4t/) + T 2 s cos (3a; + 4y). 
Hence z = T V sin (3x + ±y)+-h cos ( 3a; + 4 2/) + 2J^ e te + A 'y, 
where h 2 -hk-2h=0. 

Examples for solution. 

(1) (D- D' -l)(D- D' -2)z = e**-v. 

(2) s + p-q = z + xy. (3) (D- D' 2 )z = cos (x-3y). 

(4) r-s+p-l. (5) g-g^-f-^*. 

(6) (Z)-3Z)'-2) 2 z = 2e 2 *tan(2/ + 3a;). 

152. Examples in elimination. We shall now consider the result 
of eliminating an arbitrary function from a partial differential 
equation of the first order. 

Ex. (i). 2px-qy = <p(x 2 y). 

Differentiating partially, first with respect to x and then toy/, we get 
2rx -sy + 2p = 2xyqy'(x 2 y), 
and 2sx-ty-q = x 2 (/)'(x 2 y), 

whence x(2rx -sy + 2p) = 2?/(2s:r -ty-q) 

or 2x 2 r - 5xys + 2y 2 t + 2 {px + qy) = 0, 

which is of the first degree in r, s, t. 



180 DIFFERENTIAL EQUATIONS 

The same equation results from eliminating \fs from 
px-2qy = \fs(xy 2 ). 

Ex. (ii). p 2 +q = <j>{2x + y). 

This gives 2pr + s = 2<p'{ 2x + y), 

and 2ps + t = <p'(2x + y), 

whence 2pr + s = ips + 2t, 

again of the first degree in r, s, t. 

Ex. (iii). y-p = <p(x-q). 

This gives -r = (l -s)cp'(x-q), 

and - l-s= -t<f>'(x-q), 

whence rt = (l-s) 2 

or 2s + (rt-s 2 ) = l. 

This example differs from the other two in that p and q occur in j 
the arbitrary function as well as elsewhere. The result contains a 
term in (rt - s 2 ) 

Examples for solution. 

Eliminate the arbitrary function from the following : 

(1) py-q + 3y 2 = (j>(2x + y 2 ). (2) x-- = q>(z). 

(3) p + x-y = <p(q-2x + y). (4) px + qy = (f>(p 2 + q 2 ). 

(5) p 2 -x = <p(q 2 -2y). (6) p+zq = <p(z). 

153. Generalisation of the preceding results. If u and v are 
known functions of x, y, z, p, q, and we treat the equation u = (J> (v) 
as before, we get 

du du du du ( dv dv dv dv\ .. . 

, du du du du ( dv dv dv dv\ .. . 

and ^V^a^l^VaT/^l)-^"'- 

Eliminating <j)'(v) we find that the terms in rs and st cancel out, 

leaving a result of the form 

Rr+Ss+Tt + U(rt-s 2 ) = V, 

where R, S, T, U and V involve p, q, and the partial differential 

coefficients of u and v with respect to x, y, z, p, q. 

„ du dv dv du 
The coefficient u = =- _ — , - ~- , 

dp dq dp oq 

which vanishes if v is a function of x, y, z only and not of p or q. 

These results will show us what to expect when we start with 

the equations of the second order and try to obtain equations of the 

first order from them. 



SECOND AND HIGHER ORDERS 181 

154. Monge's method of integrating Rr + Ss+Tt=V. We shall 
now consider equations of the first degree in r, s, t, whose coefficients 
R, S, T, V are functions of p, q, x, y, z, and try to reverse the process 
of Arts. 152 and 153. 

Since dp = * dx + ^ dy = rdx+sdy 

and dq=sdx+tdy, 

Rr+Ss+Tt-V=0 

becomes *(&-"£*) + Ss + t(*^) -7-0, 

i.e. Rdpdy + T dqdx -V dy dx -s (R dy 2 -S dy dx + T dx 2 ) =0. 

The chief feature of Monge's method is obtaining one or two 
relations between p, q, x, y, z (each relation involving an arbitrary 
function) to satisfy the simultaneous equations 
Rdy 2 -Sdydx + Tdx 2 =0, 
Rdpdy + T dqdx- V dy dx =0. 

These relations are called Intermediate Integrals. 
The method of procedure will be best understood by studying 
worked examples. 

Ex. (i) . 2x 2 r - 5xys + 2yH + 2 (px + qy) = 0. 

Proceeding as above, we obtain the simultaneous equations 

2x 2 dy 2 + 5xydydx + 2y 2 dx 2 =0, (1) 

and 2x 2 dpdy + 2y 2 dqdx + 2{px + qy)dydx = (2) 

(1 ) gives (x dy + 2y dx) (2xdy + y dx) = 0, 

i.e. x 2 y — a or xy 2 = b. 
If we take x 2 y = a and divide each term of (2) by xdy or its equivalent 
-2ydx, we get 2xdp-ydq + 2pdx-qdy=0, 

i.e. 2px — qy = c. 
This, in conjunction with x 2 y = a, suggests the intermediate integral 
2px-qy = <f>(x 2 tj), (3) 

where <j> is an arbitrary function. [Cf. Ex. (i) of Art. 152.] 
Similarly xy 2 = b and equation (2) leads to 

px-2qy = \p-{xy 2 ) (4) 

Solving (3) and (4), 

3px = 2<p{x 2 y)-\l,{xy 2 ), 

3qy = (t>{x 2 y)-2\Js{xy 2 ), 



182 DIFFERENTIAL EQUATIONS 

so dz = pdx + qdy = i<l>(x ? y).(^ + ^)-^(xy 2 ).^ + ~^y 

i.e. z = ^<p {xhj) . d log {x 2 y) - ij ^ (xy 2 ) . d log {xy 2 ), 
or z=f(x 2 y) + F{xy 2 ). 

Ex. (ii). y 2 r-2ys + t=p + 6y. 

Eliminating r and t as before, we are led to the simultaneous equa- 
tions ij 2 dy 2 + 2ydydx + dx 2 =0, (5) 

and y 2 dp dy + dq dx - (p + 6y) dy dx =0 (6) 

(5) gives {ydy + dx) 2 =0, 

i.e. 2x + y 2 = a. 
Using this integral and dividing each term of (6) by y dy or its 
equivalent - dx, we get 

ydp-dq + (p + 6y)dy=0, 
i.e. py - q + Sy 2 = c. 
This suggests the intermediate integral 

py-q+3y 2 = <p(2x + y 2 ). 
As we have only one intermediate integral, we must integrate this 
by Lagrange's method. 

The subsidiary equations are 

dx dy dz 

i; = ~l =: -3y 2 + <t>(2x + y 2 )' 
One integral is 2x + y 2 = a. Using this to find another, 
dz + {-3y 2 + <j>(a)}dy = 0, 
i.e. z-y z + y<j>{2x + y 2 )=b. 
Hence the general integral is 

yf,{z-f + y<p(2x + y 2 ), 2x + y 2 }=0, 
or z — y*-y<f>(2x + y 2 )+f(2x + y 2 ). 

Ex (iii). pt-qs^q*. 

The simultaneous equations are 

qdydx+pdx 2 =0, (7) 

and pdqdx-q 3 dydx=0 (8) 

(7) gives dx=0 or qdy + pdx(=dz)—0, 

i.e. x = a or z = b. 
If dx = (8) reduces to 0=0. 
If z = b, qdy= -pdx and (8) reduces to 

pdq + q 2 p dx = 0, 
i.e. dq/q 2 + dx=0, 

giving -- + x = c = \fr(z) (9) 



SECOND AND HIGHER ORDERS 183 

(9) may be integrated by Lagrange's method, but a shorter way is 
to rewrite it fi v \ 

giving y = xz- I \fs(z)dz + F(x) 

y = xz+f(z) + F(x). 
Examples for solution. 

(1) r — l cos 2 x + p tan x=0. 

(2) (x - y) {xr -xs-ys + yt) =(x + y) (p - q). 

(3) (q + l)s = (p + l)t. (4) t-rscc*y = 2qtany. 

(5) xy(t-r) + (x 2 -y 2 )(s-2)=py- qx. 

(6) (l+q) 2 r-2(l+p + q + pq)s + (l+p) 2 t=0. 

(7) Find a surface satisfying 2x 2 r-5xys + 2yH + 2(px + qy) =0 and 
touching the hyperbolic paraboloid z = x 2 -y 2 along its section by the 
plane y = 1 . 

(8) Obtain the integral of q 2 r-2pqs+pH = in the form 

y + xf(z) = F(z), 
and show that this represents a surface generated by straight lines that 
are all parallel to a fixed plane. 

♦155. Monge's method of integrating Rr +Ss +Tt +U(rt-s 2 ) V. 
As before, the coefficients R, S, T, U, V are functions of p, q, 
x, y, z. 

The process of solution falls naturally into two parts : 
(i) the formation of intermediate integrals ; . 
(ii) the further integration of these integrals. 
For the sake of clearness we shall consider these two parts 
separately. 

156. Formation of intermediate integrals. As in Art. 154, 
r = {dp - s dy)/dx 
and t = (dq-s dx)/dy. 

Substitute for r and &'m 

Rr+Ss + Tt + U{rt-s 2 ) = V, 
multiply up by dx and dy (to clear of fractions), and we get 
R dp dy + Tdqdx + U dp dq - V dx dy 
-s(R dy 2 -Sdxdy + Tdx 2 + U dp dx + U dq dy) = 0, 
say M -sN=0. 

* The remainder of this chapter should be omitted on a first reading. This 
extension of Monge's ideas is due to Andre Marie Ampere, of Lyons (1775- 1830), 
whose name has been given to the unit of electric current. 



184 DIFFERENTIAL EQUATIONS 

We now try to obtain solutions of the simultaneous equations 

M=0, 
N=0. 
So far we have imitated the methods of the last paragraph, but 
we cannot now factorise N as we did before, on account of the 
presence of the terms U dpdx + U dq dy. 

As there is no hope of factorising M or N separately, let us try 
to factorise M+XN, where X is some multiplier to be determined 
later. 

Writing M and N in full, the expression to be factorised is 
Rdy 2 +T dx 2 -{S +XV)dxdy + U dpdx + U dq dy 
+ XRdp dy + XT dq dx+XU dp dq. 
As there are no terms in dp 2 or dq 2 , dp can only appear in one 
factor and dq in the other. 
Suppose the factors are 

Ady+Bdx+C dp and Edy+Fdx+Gdq. 
Then equating coefficients of dy 2 , dx 2 , dp dq, 

AE = R; BF = T; CG=XU. 
We may take 

A=R, E = l, B = kT, F = l/k, C = mU, G = X/m. 
Equating the coefficients of the other five terms, we get 

JcT+R/k= -(S+XV), (1) 

XR/m = U, (2) 

JcTX/m=XT, (3) 

mU=XR, (4) 

mU/k = U (5) 

From (5), m = k, and this satisfies (3). 
From (2) or (4), m = XR/U. 
Hence, from (1), 

X 2 (RT + UV)+XUS + U 2 =0 (6) 

So if A is a root of (6), the factors required are 

(Rdy+X y-j dx +XR dp) (dy + w-, dx + p dq), 

7? 1 

i.e. j-j (U dy +XT dx +XU dp) . ^ Q^R dy + U dx+XUdq). 

We shall therefore try to obtain integrals from the linear 

equations U dy +XT dx +XU dp =0 (7) 

and XRdy + Udx+XUdq=0, (8) 

where X satisfies (6). 



SECOND AND HIGHER ORDERS 185 

The rest of the procedure will be best understood from worked 
examples. 

157. Examples. 

Ex. (i). 2s + (rt-s*)=*l. 

Substituting R=T = 0, S=2, £/=F=l in equation (6) of the last 
article,* we get \2 + 2\ + 1 = 0, 

a quadratic with equal roots -1 and -1. 
With X= -1, equations (7) and (8) give 

dy — dp = 0, 
dx-dq — 0, 
of which obvious integrals are 

y — p = const, 
and x - q = const. 

Combining these as in Art. 154, we get the intermediate integral 
y-p=f{x-q). 

Ex. (ii). r + 3s*+t + {rt-s 2 )=>l. 

The quadratic in X comes to 

2X 2 + 3X + 1=0, 
so X = - 1 or - 1 . 

With X= -1, equations (7) and (8) give 

dy-dx-dp=0, 
-dy + dx-dq =0, 
of which obvious integrals are 

p + £-?/= const (1) 

and q-x + y = const (2) 

Similarly X = - J leads to 

p + x-2y = comt. (3) 

and 9 -2a; + y= const (4) 

In what pairs shall we combine these four integrals ? 

Consider again the simultaneous equations denoted by M=0, N = 
in the last article. If these are both satisfied, then M + X 1 iV = and 
M + \ 2 N =0 are also both satisfied fwhere X x and X 2 are the roots of the 
quadratic in X). Therefore one of the linear factors vanishes for X = X, 
and one (obviously the other one, or else dy = 0) for X = X 2 . 

That is, we combine integrals (1) and (4), and also (2) and (3), 
giving the two intermediate integrals 

p + x-y=f(q-2x + y) 
and p + x - 2y = F(q - x + y). 

* We quote the results of the last article to save space, but the student is 
advised to work each example from first principles. 



186 DIFFERENTIAL EQUATIONS 

Ex. (iii). 2yr + (px + qy)s + xt- xy(rt - s 2 ) = 2 - pq. 
The quadratic in X comes to - 

X 2 xypq - \xy (px + qy)+ x 2 y 2 = 0, 
giving X = y/P or x/q. 

Substituting in (7) and (8) of the last article, we get, after a little 

reduction, pdy-dx + ydp=Q, (5) 

2y dy - px dx - xy dq=0, (6) 

- qy dy + x dx - xy dp = 0, (7) 

and -2dy + qdx + xdq=0 (8) 

Combining the obvious integrals of (5) and (8), we get 
yp-x=f(-2y + qx). 

But (6) and (7) are non-integrable. This may be seen from the 
way that p and q occur in them. Thus, although the quadratic in X has 
two different roots, we get only one intermediate integral. 

Examples for solution. 

Obtain an intermediate integral (or two if possible) of the following : 
(1) 3r + is + t + (rt-s 2 )=l. (2) r + t-(rt-s 2 )=l. 

(3) 2r + te x -(rt-s 2 )=2e x . (4) rt-s 2 + 1=0. 

(5) 3s + (rt-s 2 ) =2. 

(6) qxr + (x + y) s + pyt + xy (rt - s 2 ) = 1 - pq. 

(7) (q 2 - 1) zr - 2pqzs + (p 2 -l)zt + z 2 (rt - s 2 ) = p 2 + q 2 - 1 . 

158. Further integration of intermediate integrals. 

Ex. (i). Consider the intermediate integral obtained in Ex. (i) of 
Art. 157, y_2i=f(x-q). 

We can obtain a " complete " integral involving arbitrary constants 
a, b, c by putting x-q = a 

and y - p =f (a) = b, say. 

Hence dz =p dx + q dy = (y -b) dx + (x- a) dy 

and z=xy- bx - ay + c. 

An integral of a more general form can be obtained by supposing 
the arbitrary function / occurring in the intermediate integral to be 
linear, giving y _ p = m ( x - q )+ n . 

Integrating this by Lagrange's method, we get 
z = xy + <p(y + mx) - nx. 

Ex. (ii). Consider the two intermediate integrals of Ex. (ii), Art. 157, 
p + x-y=f(q-2x + y) 
and p + x-2y = F(q-x + y). 



SECOND AND HIGHER ORDERS 187 

If we attempt to deal with these simultaneous equations as we dealt 
with the single equation in Ex. (i), we get 

q - 2x + y = a, 
q-x+y=fi, 
p + x-y=f(a), 
p + x-2y = F(j3). 
If the terms on the right-hand side are constants, we get the absurd 
result that x, y, p, q are all constants ! 

But now suppose that a and /3 are not constants, but parameters, 
capable of variation. 

Solving the four equations, we get 
x = (5-a, 
V-f(a)-F(fa 
p = y-x+f(a), 
q=, x -y + (3, 
giving dz=pdx + qdy 

= (y-x) (dx - dy) +f (a) dx + (3 dy 

= - \d {x - y)* +/(a) d(3 -/(a) da + fif'(a) da - fiFtf) d/3 : 

i.e. z =-l(x-y)2-^f(a)da-^F'(p)dp + Pf(a). 

To obtain a result free from symbols of integration, put 

( f(a)da = <j>(a) and f F(/3) rf/3 = ^ ((3). 

Now [ @F'(p) d/3 - f3F{(3) - \F(/3) d/3, integrating by parts, 

= /3V/(/3)-V'(/3) 
Hence z = - £ (x - yf - <f> (a) - /3y/(/3) + ^ (/3) + /3<p'(a), 

rz=-h(x-y) 2 -<p(a) + ^( l 8)+(3y, 
or finally j x = /3 - a, 

These three equations constitute the parametric form of the equation 
of a surface. As the solution contains two arbitrary functions, it may 
be regarded as of the most general form possible. 

Examples for solution (completing the solution of the preceding set). 
Integrate by the methods explained above : 
(1) p + x-2y=f{q-2x + 3y). (2) p-x=f(q-y). 

(3) p-e*=f{q-2y). (4) p-y=f(q + x), 

p + y = F(q-x). 

(5) P-V =/ (q - 2s), (6) px - y =f (qy - x). 

p-2y = F(q-x). (7) (zp-x)=f(zq-y). 

(8) Obtain a particular solution of (4) by putting 0(a)=-£a 2 , 
\fr ((3) = 1/3 2 and eliminating a and (3. 



188 DIFFERENTIAL EQUATIONS 

MISCELLANEOUS EXAMPLES ON CHAPTER XIV. 

(1) r = 2y 2 . (2) log s = x + y. (3) 2yq+y*t = l. 

(4) r-2s + t = sm{2x + 3y). (5) x 2 r-2xs + t + q=0. 

(6) rx 2 -3sxy + 2ty 2 +px + 2qy = x + 2y. 

(7) y 2 r + 2xys + x 2 t + px + qy = 0. 

(8) 5r + Gs + 3t + 2(rt-s 2 )+3 = 0. 

(9) 2pr + 2ql-4:pg(rt-s 2 ) = l. 

(10) rt - s 2 - s (sin a; + sin y) = sin x sin */. 

(11) 7r-8s-3i + (rt-s 2 )=36. 

(12) Find a surface satisfying r = 6x + 2 and touching z = x 3 + y 3 
along its section by the plane x + y + 1 = 0. 

(13) Find a surface satisfying r-2s + t = 6 and touching the hyper- 
bolic paraboloid z = xy along its section by the plane y = x. 

(14) A surface is drawn satisfying r + l = and touching x 2 + z 2 — l 
along its section by y=0. Obtain its equation in the form 

z 2 (x 2 + z 2 -l) = y 2 (x 2 + z 2 ). [London.] 

(15) Show that of the four linear differential equations in x, y, p, q 
obtained by the application of Monge's method to 

2r + qs + xt - x (rt - s 2 ) = 2, 
two are integrable, leading to the intermediate integral 

P-x=f(qx-2y), 
while the other two, although non-integrable singly, can be combined 
to give the integral ? , + J ? 2 _ x = a> 

Hence obtain the solutions 

z = \x 2 - 2mxy - f mPx 3 + nx + <p(y + hnx 2 ) 
and z = (a- \h 2 )x + \x 2 + by + c, 

and show that one is a particular case of the other. 

(16) A surface is such that its section by any plane parallel to x=0 
is a circle passing through the axis of x. Prove that it satisfies the 
functional and differential equations 

y 2 + z 2 + yf(x)+zF(x)=0, 

(y 2 + z 2 )l + 2(z-yq)(l+q 2 )=0. 

(17) Obtain the solution of x 2 r + 2xys + y 2 t = in the form 

*=/(!)^(f> 

and show that this represents a surface generated by lines that intersect 
the axis of z. 

(18) Show that rt-s 2 = leads to the " complete " integral 

z = ax + by + c. 






MISCELLANEOUS EXAMPLES 189 

Show that the " general " integral derived from this (as in Art. 134) 
represents a developable surface (see Smith's Solid Geometry, Arts. 
222-223). 

Hence show that for any developable surface q=f(p). 

(19) Find the developable surfaces that satisfy 

pq(r -t)- (p 2 -q 2 )s + {py - qx) (rt - s 2 ) =0. 
[Assume q =/(/>). This is called Poisson's method. We get 
q = ap or p 2 + q 2 = b 2 , 
giving z = <p(x + ay) or z = bx cos a + by sin a + c. 

The second of these integrals represents a plane which generates the 
developable surface given by the corresponding " general " integral.] 

(20) Show that if 

X=p, Y = q, Z=px + qy-z, 

then r=TJ(RT-S 2 ), s= -S/(RT-S 2 ), t = R/(RT-S 2 ), 
where R = ^y^, etc. 

Hence show that the equation 

ar + bs + ct + e (rt - s 2 ) = 
transforms into AT - BS + CR + E = 0, 

where a, b, c, e are any functions of x, y, p, q, and A, B, C, E the corre- 
sponding functions of P, Q, X, Y. 

Apply this Principle of Duality (cf. No. 21 of the Miscellaneous 
Examples at the end of Chap. XII.) to derive two intermediate integrals 
°f pq(r -t)- (p 2 -q 2 )s + (py - qx) (rt - s 2 ) =0. 

(21) Prove that if x, y, u, v are real and u + iv=f(x + iy), then F = ?t 
and V — v are both solutions of 

d 2 V dW 
dx 2 dy 2 ' 
and the two systems of curves a = const., 

v= const., 
are mutually orthogonal. 

Verify these properties for the particular cases 
(i) u + iv = x + iy, 
(ii) u + iv = (x + iy) 2 , 
(iii) u + iv = ]((x + iy). 
[The differential equation is the two-dimensional form of La place's 
equation, which is of fundamental importance in gravitation, electro- 
statics and hydrodynamics, u and v are called Conjugate Functions. 
See Ramsey's Hydro-Mechanics, Vol. II. Art. 11. J 

(22) Obtain the solution of 

a 2 // J^y 
dt 2 """dx 2 ' 



190 DIFFERENTIAL EQUATIONS 

subject to the conditions y=f(x) and ~ = F(x) when «=0, in the form 

1 Cx+at 

y = U(x + at) + $f(x-at) + -\ F(\)d\. 

taJx-at 

[y is the transverse displacement of any point x of a vibrating 
string of infinite length, whose initial displacement and velocity are 
given by f(x) and F(x). See Ramsey's Hydro- Mechanics, Vol. II. 
Art. 248.] 

(23) If y=f(x) cos (nt + a) is a solution of 

dt 2 +(l d*4~ U ' 

show that f(x)=A sin mx + B cos mx + H sinh mx + K cosh mx, where 

m = \/(n/a 2 ). 
[The differential equation is that approximately satisfied by the 
lateral vibrations of bars, neglecting rotatory inertia. See Rayleigh's 
Sound, Art. 163.] 

(24) Show that 

w = A sin (mirx/a) sin (mry/b) cos {pet + a) 
,. a d 2 w vfdhv d 2 w\ 

satafies W=°\w + W>' 

and vanishes when 

a:=0, y = 0, x = a or y = b, 
provided that m and n are positive integers satisfying 
(p/7r) 2 = (m/a) 2 + (w/6) 2 . 
[This *gives one solution of the differential equation of a vibrating 
membrane with a fixed rectangular boundary. See Rayleigh's Sound, 
Arts. 194-199.] 

(25) Show that w=AJ {nr) cos (nct + a) 
,. o d 2 w /d 2 w 1 dw 

Satlsfi6S " Tt 2 ^ C \ji 2+ rTr 

where J is Bessel's function of order zero (see Ex. 2 of the set following 

Art. 97). 

[This refers to a vibrating membrane with a fixed circular boundary. 
See Rayleigh's Sound, Arts. 200-206.] 

(26) Show that V = (Ar n + Br-"- 1 ) P n (cos 6) 

. _ a 2 F 2 3F i dw cot oar rt 

where P„ is Legendre's function of order w (for Legendre's equation, 
see Ex. 2 of the set following Art. 99). 

[N. B. — Take /j. = cos 6 as a new variable. This equation is the 
form taken by Laplace's potential equation in three dimensions, when 
V is known to be symmetrical about an axis. See Routh's Analytical 
Statics, Vol. II. Art! 300.] 



APPENDIX A 

The necessary and sufficient condition that the equation M dx + N dy = 

should be exact 

(a) If the equation is exact, 

M dx + N dy = a, perfect differential = df, say. 

So M = % and N = % > 

- dx oy 

it . dN a 2 / d 2 f bm 

therefore — = — -;- = — i- = — - , 

ox oxoy oyox oy 

so the condition is necessary. 

(6) Conversely, if — = — , put F=\Mdx, where the integration 

is performed on the supposition that y is constant. 

„, dF __ d 2 F d 2 F dM BN 

Ihen -=- = M and - - = =—=- = -=- = -=- . 

ox oxoy oyox oy ox 

* £(*-$-* 

N - -=— = a constant as far as a; is concerned, that is, 
J a function of y, 

= 0(y)»say. 
Then iV = ^ + 0(y). 

Now put /= F + I («/) %. 

Then N = %£. 

oy 

dF 
Also M = -=- by definition of F 
ox 

= 4~, since F and /differ only by a function of y. 

Thus M dx + N dy = ~- dx + ~- dy = df, a perfect differential. 

So the equation is exact, that is, the condition is sufficient. 

d 2 f d 2 f ~~~ * 

[Our assumption that =— ■ ~ = ~— "i is justified if / and its first and 

second partial differential coefficients are continuous. See Lamb's 
Infinitesimal Calculus, 2nd ed., Art. 210.] 

191 



APPENDIX B 

The equation P(x, y, z) ~- +Q(x, y, z) —- + R(x, y, z)~ = 0, regarded as 
four-dimensional, has no special integrals. (See Art. 127.) 
Let u(x, y, z)=a, 

v(x, y, z) = b, 
be any two independent integrals of the equations 
dx/P = dy/Q = dz/R. 
Then we easily prove that 

du Bu du _ ., . 

P S - X +Q di + R S -z = ° (1) 

and p^ +Q f_ +B 3jl _ (2) 

Bx By Bz 

The left-hand side of (1) does not contain a, and therefore cannot 
vanish merely in consequence of the relation u = a. Hence it must 
vanish identically. Similarly equation (2) is satisfied identically. 

Now let f=w(x, y, z) be any integral of the original partial 
differential equation, so that 

n Biv „Bw n Bw _ /0 , 

P 8^ +Q S l j +R di =<> (3) 

This is another identical equation, since / does not occur in it. 
Eliminating P, Q, R from (1), (2), (3), we get 

^4=0 identically. 

B(x,y,z) 

Hence w is a function of u and v, say 

W = (f)(u, v). 

That is, /= w is part of the General Integral, and therefore, as /= w 
is any integral, there are no Special Integrals. 

[The student will notice the importance in the above work of a 
differential equation being satisfied identically. Hill's new classification 
of the integrals of Lagrange's linear equation (Proc. London Math. Soc. 
1917) draws a sharp distinction between integrals that satisfy an 
equation identically and those which have not this property.] 



192 



APPENDIX 

The expression obtained for dz by Jacobi's method of solving a single 
partial differential equation of the first order (Art. 140) is always 
integrable. 

To prove that dz = p x dx x + p 2 dx 2 + p 3 dx 3 

is integrable it is necessary and sufficient to prove that 

L = M = N=0, (A) 

where iJ^-^, mJ^-^, nJ^-^. 

OX 3 OX 2 OXy ox 3 ax 2 ox l 

Now, by adding equations (8), (9), (10) of Art. 140 and using the 
relation (F, F l ) = 0, but not assuming the truth of (A), we get 

I'gilA +M Wfi +s 'JL3i =0 (B, 

o(Pt, Ps) o(p 3 , p x ) d(p v p 2 ) 

a . ., . T d{F v F 2 ) ..d(F v F i ) . T d(F v F 2 ) _ 

Similarly L -^—^ — ^ + M^-^ — % + N-±-J> — ^=0 (C) 

oiPvto) °iPz>Pi) d(p v p 2 ) 

T d(F 2) F) .,d(F 2 ,F) „d(F 2 ,F) n 

and £ a ; 8 ; +M ~- 2 — [ +N~^ — =0 (D) 

o(p 2 , Pz) o{p 3 , p x ) d(p lt p 2 ) 

From equations (B), (C), (D) we see that either L — M=N—0 or 
A = 0, where A is the determinant whose constituents are the 
coefficients of L, M, N in (B), (C), (D). 

But these coefficients are themselves the co- factors of the constituents 
of the determinant a/ ™ w p \ 

~d(Pi>P2>Ps)' 
and by the theory of determinants A=«7 2 - 

Now J cannot vanish,* for this would imply the existence of a 
functional relation which would contradict the hypothesis of Art. 140 
that the p's can be found as functions of the as's from 

F=F 1 -a i = F 2 -a 2 = 0. 
Hence A^0; therefore L = M = N = 0. 

* All the equations of this appendix are satisfied identically. 



193 



APPENDIX D 

Suggestions for further reading 

No attempt will be made here to give a complete list of works on 
differential equations. We shall merely give the names of a very 
small number of the most prominent, classified in three sections. 

I. Chiefly -of analytical interest (forming a continuation to Chapter X.). 

(a) Forsyth : Theory of Differential Equations (1890 and later years, 
Cambridge Univ. Press). 

This important work is in six volumes, and is the most exhaustive 
treatise in English upon the subject. It should not be confused with 
his more elementary work in one volume (4th ed. 1914, Macmillan). 

(b) Goursat : Cours d' Analyse matMmatiqne, Vols. II. and III. (2nd 
ed. 1911-15, Gauthier-Villars ; English translation published by Ginn). 

This deals almost entirely with existence theorems. 

(c) Schlesinger : Handbuch der Theorie der linearen Differential- 
gleichungen (1895-8, 3 vols, Teubner). 

II. Partly analytical but also of geometrical interest. 

(a) Goursat : Equations aux dirivies partielles du premier ordre (1891). 

(b) Goursat : Equations anx de'rive'es partielles du second ordre 
(1896-98, 2 vols., Hermann et fils). 

(c) Page : Ordinary differential equations from the standpoint of Lie's 
Transformation Groups (1897, Macmillan). 

This deals with the elements of differential equations in a highly 
original manner. 

III. Of physical interest (forming a continuation to Chapters III. and IV.). 

(a) Biemann : Partielle Differentialgleichungen und deren Anwendung 
auf physikalische Fragen (1869, Vieweg). 

(b) Riemann- Weber : A revised edition of (a), with extensive 
additions (1900-01, Vieweg). 

(c) Bateman : Differential Equations (1918, Longmans). 
This contains many references to recent researches. 

It is impossible to mention original papers in any detail, but the 
recent series of memoirs by Prof. M. J. M. Hill in the Proceedings of tlie 
London Mathematical Society should not be overlooked. 



194 



MISCELLANEOUS EXAMPLES ON THE WHOLE BOOK 

. dy ^y* + 2,x 2 y 

y () dx~a? + 3xyf [London.] 

? (2) £ + ^V = 2« (1 + a 2 ). [London. ] 

>(3) tan ?/ ^ + tan a; = cos y cos 3 z. [London.] 

(4) ^ 2 *i + (!) 2 London.] 

(5) (1 -x 2 )-£-xy = x 2 y 2 . [London.] 

(6) (Z> 2 + 4)*/ = sin2z. [London.] 

(7) (D 3 -D 2 + 3D + 5)y = x 2 + e*cos2x. [London.] 

(8) (x 3 D 3 + x 2 D 2 )y = l+x + x 2 . [London.] 

/«» • dy 

(9) cos x sin cc -— = y + cos x. [London.] 



[London.] 



(10) ^- = » + y + 2cos«,- 

dt~ 6x y - 

(11) ^/ = a; (^) 3 + l- [London.] 

(13) (D 4 + 8D 2 + 16)?/ = a;cos2a:. [London.] 

(14) I x 2 dy+ I xydx = x 3 . [London.] 

(15) (?/ 2 + yz - z) dx + (x 2 + xz - z) dy + (x + y- xy) dz = 0. [London . ] 

(16) (2x* - y 3 - z 3 ) yz dx + (2y 3 - z 3 - x 3 ) zx dy + (2z 3 - x 3 - y 3 ) xy dz = 0. 

[London.] 

(17) xp-yq + (x 2 -y 2 )=0. [London.] 

(18) (x + 2y -z)p + (3y-z)q = x + y. [London.] 

195 



.2 



196 DIFFERENTIAL EQUATIONS 

/■.^v 2 (xz - yz + xy) _ _ , , 

(19) xp + yq+ \ 3 X + Z =0 ' London.] 

(20) p(x + p)+q(y + q) = z. [London.] 

(21) r + s=p. [London.] 

(22) z-\px-qy=p 2 \x 2 . [London.] 

(23) r-x = t-y. [London.] 

(24) z = px + qy-sxy. [London.] 

(25) z \rt - s 2 ) + pqs = 0. [London. ] 

(26) x 2 r + 2xys + y 2 t = xy. [London.] 

(27) rq(q + l)-s(2pq+p + q + l)+tp(p + l)=0. [London.] 

(28) y 3 =xy 2 p + x*p 2 . [Math. Trip.] 

(30) Pi - - i?- + x 2n y=0. [Math. Trip.] 
ax 2 x ax 

(31) (zp + x) 2 + (zq + ij) 2 = l. . [Math. Trip.] 

(f 2 y dy 

(32) Find a solution of the equation j\- 3 ■— + 2y = e 3x which shall 

vanish when x=0 and also when x = log e 2. [Math. Trip.] 

(33) Solve the equation 

d oc doc 

-j-£ + 2k-j-+ (k 2 + \ 2 )x = A cos pt. 

Show that, for different values of p, the amplitude of the particular 
integral is greatest when p 2 = \ 2 -K 2 , and prove that the particular 
integral is then 

(A/2k\) cos (pt - a), where tan a =p/k. [London.] 

(34) Solve the equation 

d 2 v dy o 

T^ + ~r tanaj + tycos-x = 
dx l ax 

by putting z = sin x. 

d 2 V d 2 V d 2 V 

(35) (i) Assuming a solution of ■=—, r + -r— r + -=-=- = to be of the 

ox 1 ay oz* 

form F(r + z), where r 2 = x 2 + y 2 + z 2 , obtain the function F ; and by 

integrating with respect to z, deduce the solution V = z\og (r + z) -r. 

dV d 2 V 
(ii) Assuming a solution of _- = a 2 2 to be of the form <f> {£), 

where i=x/^/t, obtain the function <f>; and deduce a second solution 
by differentiating with respect to ,r. [London.] 

(36) Obtain a rational integral function V of x, y, z which satisfies 
the condition 92 j/ 92 y $iy 

dx 2 dy 2 dz 2 ' 
and is such as to have the value A: 4 at points on the surface of a sphere 
of unit radius with its centre at the origin. [Math. Trip.] 



MISCELLANEOUS EXAMPLES 197 

(37) Show that a solution of Laplace's equation V 2 m = is 

u = (A cos 116 + B sin nO) e-* 2 J n (\r), 
where r, 6, z are cylindrical co-ordinates and A, B, n, \ are arbitrary 
constants. [London.] 

(38) Show that J n (r) (a n coa n$ + b n sin nO), where r and 6 are 
polar co-ordinates and a n and b n are arbitrary constants, is a solution 
of the equation d 2 V d 2 V 

fa* + "^2 + V = °- [London.] 

(39) Show how to find solutions in series of the equation 

du_ 2 d 2 u 

& =a dx 2 ' 

and solve completely for the case in which, when x = 0, 

du _, , _ , , 

w = a -=- = cosh £. [London .] 

(40) Obtain two independent solutions in ascending powers of x of 
the equation ^2 ? , 

and prove by transforming the variables in the equation, or otherwise 
that the complete solution may be written in the form 

y = Ax h J^ (a; 1 ) + Bx*J_ h (x*), 

where A and B are arbitrary constants. [London.] 

(41) Show that the complete solution of the equation 

C ^ + P + Qy + Ry 2 = 0, R.^-rr. 

where P, Q, R are functions of x, can be obtained by the substitution 
y = yi + l/z, if a particular solution, y lt is known. 

Show that, if two particular solutions y x and y 2 are known, the 
complete solution is 

lo s \^~i7 ) = \ R (y» - yi) dx + const - 

Obtain the complete solution of the equation 

(x 2 -l) C ^ + x + l-(x 2 + l)y + (x-l)y 2 = 0, 

which has two particular solutions, the product of which is unity. 

..„. _,. , , ,.«■ • , • [London.] 

(42) Show that the differential equation 

(l- x 2)^ 2 + 2{b + (a-l)x}^ x + 2ay=0 

has a solution of the form (1 +x) p (l - x)i, where p and q are determinate 
constants. Solve the equation completely ; and deduce, or prove 
otherwise, that if 2a is a positive integer n, one solution of the equation 
is a polynomial in x of degree n. [London.] 



198 DIFFERENTIAL EQUATIONS 

(43) Verify that 1 - x 2 is a particular solution of the equation 

x (1 - x 2 ) 2 ^| + (1 - x 2 ) (1 + 3x 2 ) ^ + ix (1 + x 2 ) y =0, 

and solve it completely. 

By the method of variation of parameters or otherwise, solve com- 
pletely the equation obtained by writing (1 -a; 2 ) 3 instead of zero on the 
right-hand side of the given equation. [London.] 

(44) Show that the complete solution of the equation 

where P, Q are given functions of x, can be found if any solution of the 
equation du j d p 1 

dx 2& 4 

is known. 

Hence, or otherwise, solve the equation 

(l-x 2 )g-4^ + (z*-3) *, = (). [London.] 

(45) Prove by putting v = we x that the complete solution of the 

d 2 v dv n , 

equation x j-^ -Zn-j-+xv = 0, where n is an integer, can be expressed 

in the form 

(A cosx + B sin x)f(x) + (A sin x - B cos x) <f> (x), 

where f(x) and (p (x) are suitable polynomials. [London.] 

(46) If u, v are two independent solutions of the equation 

f^)y"'-f\x)y" + ( p{x)y' + x {x)y = 0, 

where dashes denote differentiation with regard to x, prove that the 
complete solution is Au + Bv + Cw, where 

f vf(x) dx f «f (x) dx 
J (uv -uv) 1 J (uv -uv) 2 

and A, B, C are arbitrary constants. 
Solve the equation 

x 2 (x 2 + 5)y'" -x(7x 2 + 25)y" + (22x 2 + i0)y' -30xy=0, 
which has solutions of the form x n . [London.] 

(47) Obtain two independent power-series which are solutions of 

the equation dh d 

(x 2 -a 2 )- T ~ + bx^ + cm = 0, 
ax* dx 

and determine their region of convergence. [London.] 

(48) Prove that the equation 



MISCELLANEOUS EXAMPLES 199 

has two integrals 

where aw = ( l> + l) j ' [London.] 

(49) Form the differential equation whose primitive is 



. / . co?a;\ n ( sinzX 
y = A ( sin x + - — J + B ( cos x J , 

where A, B are arbitrary constants. [London.] 

(50) Obtain the condition that the equation 

Pdx + Qdy = 
may have an integrating factor which is a function of x alone, and apply 
the result to integrate 

(3xy - 2a?/ 2 ) dx + (x 2 - 2axy) dy = 0. [London. ] 

(51) Show that the equations 

dy 2ax 2 dy „ 
«/-z/ + -5 — 2 / = 0, 
dx x*" - y*- dx 

x 2 - y 2 + 2(xy + bx 2 ) ( ^ = 0, 

have a common primitive, and find it. [London.] 

(52) Prove that any solution of the equation 

n d 2 u ^du n . 
P dx~ 2+Q dx + Ru = ° 
is an integrating factor of the equation 
d 2 d 

dtf {Pu) ~Tx {Qu) + Ru=0 > 

and conversely that any solution of the latter equation is an integrating 
factor of the former. 

Hence integrate the first of these equations completely, it being 
given that d 2 /P\ R . rT , . 

p(g) + r a [London - ] 

(53) If the equation fg[ + pjp +Qy =0, 

where P and Q are functions of x, admits of a solution 

y = A sin (nx + a), 
where A and a are arbitrary constants, find the relation which connects 
P and Q. [London.] 



d 2 ti 
(54) Solve the equation — 1-4?/ = 



2y 



dx 2 J (1-x) 2 ' 
having given that it has two integrals of the form 



a + bx . 



[London.] 



200 DIFFERENTIAL EQUATIONS 

(55) Show that the linear differential equation whose solutions are 

the squares of those of -J( + p_^ + n v== o 
dx 2 dx " 

may be written (l +2P ) (g + p| + 20y ) +2 q|-0. 

(56) Show that the total differential equation 

3a; 2 (y + z) dx + (z 2 - X s ) dy + (y 2 - x 3 ) dz = 
satisfies the conditions of integrability, and integrate it. [London.] 

(57) The operator -=- being represented by D, show that if X is a 

function of x and <p(D) a rational integral function of D, 
<t>(D)xX=x<j>(D)X + <f>'(D)X. 

Extend the result to the case in which l/(f>{D) is a rational integral 
function of D. 

Solve the differential equation 

d u 

— + 8y = 3x 2 + xe- 2x cosx. [London.] 

(68) Show that 3^ + 4x^-8^ = 

has an integral which is a polynomial in x. Deduce the general solution. 

[Sheffield.] 

(59) Show that, if in the equation Pdx + Qdy + Rdz = Q, P, Q, R 
are homogeneous functions of x, y, z of the same degree, then one variable 
can be separated from the other two, and the equation, if integrable, 
is thereby rendered exact. 

Integrate 

z 3 (x 2 dx + yHy) +z{xyz 2 + z 4 - {x 2 + y 2 ) 2 } {dx + dy) 

+ {x + y){z A - z 2 (x 2 + y 2 )-{x 2 + y 2 ) 2 } dz = 0, 
obtaining the integral in an algebraic form. [London.] 

(60) Show that, if the equation Pdx + Qdy + Rdz = is exact, it 
can be reduced to the form X du + mdv = ; where X//x is a function of 
u, v only and u = constant, v = constant are two independent solu- 
tions oi dx dy dz 

?Q_dR~dR_dP~dP_dQ' 
dz By dx dz dy dx 
Hence, or otherwise, integrate the equation 

(yz + z 2 ) dx - xz dy + xy dz = 0. [London.] 

(61) Prove that z 2 = 2xy is not included in 

x + y + V(z 2 - 2xy) =f(x + y + z 2 ), 
which is the general solution of 

{2V(z 2 - 2xy) -2x-l}zp + {l+2y- 2y/{z 2 - 2xy)}zq = x-y, 
but that it is nevertheless a solution of the equation. [Sheffield.] 






MISCELLANEOUS EXAMPLES 201 

(62) (i) Show how to reduce Biccati's equation 

^ = «o (») + a i (*) V + a 2 ( x ) t 

to a linear equation of the second order ; and hence or otherwise prove 
that the cross-ratio of any four integrals is a constant, 
(ii) Verify that \ + x tan x, \-x cot x are integrals of 

and deduce the primitive. [London.] 

(63) By solving ^= -<oy, 

dy 

~- = cox 

dt 

in the ordinary way, and eliminating t from the result, prove that the 
point (x, y) lies on a circle. 

Also prove this by adding x times the first equation to y times the 
second. 

[The equations give the velocities, resolved parallel to the axes, of 
a point which is describing a circle with angular velocity co.] 

(64) Find the orthogonal trajectories of the curves 

y 2 (a-x)= x 3 . 
Prove that they reduce to the system 

r 2 = b*(3 + cos26). [Sheffield.] 

(65) -j- = ny-mz, 

dy , 

-~- = lz- nx, 

at 

dz , 

Tt =mx-ly, 

where I, m, n are constants, prove that 

Ix + my + nz, 

x 2 + y 2 + z z , 

(t)'+(lD'+(l) 8 

are all constant. Interpret these results. 

(66) A plane curve is such that the area of the triangle PNT is 
m times the area of the segment APN, where PN is the ordinate, NT 
the subtangent at any point P, and A the origin ; show that its equation 
is y 2m - 1 = a 2m - 2 a;. 

Show that the volume described by the revolution of the segment 
APN about the axis of x bears a constant ratio to the volume of the 
cone generated by the revolution of the triangle PNT. [London.] 



202 DIFFERENTIAL EQUATIONS 

(67) By using the substitutions x = rcos0, y = r sin 0, or otherwise, 
solve the differential equation 

(x 2 + y 2 ) (xp - y) 2 = 1 + p 2 . 
Also find the singular solution, and interpret the results geo- 
metrically. [London.] 

(68) Show that the equation 

(x 2 + y 2 - 2xpy) 2 = ia 2 y 2 ( 1 - p 2 ) 
can be reduced to Clairaut's form by making y 2 - x 2 a new dependent 
variable ; solve it and show that the singular solution represents two 
rectangular hyperbolas. Verify also that this solution satisfies the 
given equation. [London.] 

(69) Prove that the curves in which the radius of curvature is equal 
to the length intercepted on the normal by a fixed straight line are 
either circles or catenaries. [London.] 

(70) Solve the equation 

y = x- lap + ap 2 , 
and find the singular solution, giving a diagram. [London.] 

(71) A plane curve is such that its radius of curvature p is con- 
nected with the intercept v on the normal between the curve and the 
axis of x, by the relation pv = c 2 . Show that, if the concavity of the 
curve is turned away from the axis of x, 

y 2 = c 2 sin 2 <p + b, 
where <p is the inclination of the tangent to Cx Obtain the value of 
x as a function of (p in the case &=0; and sketch the shape of the 
curve. [London.] 

(72) Show that, if the differential equation of a family of curves be 
given in bipolar co-ordinates r, /, 6, 6', the differential equation of the 
orthogonal trajectories is found by writing rclQ for dr, r' dO' for dr', 
- dr for r d0, - dr' for r'dQ'. 

Find the orthogonal trajectories of the curves 

a b 
- + -, = o, 
r r 

c being the variable parameter. [London.] 

(73) The normal at a point P of a curve meets a fixed straight line 
at the point G, and the locus of the middle point of PG is a straight 
line inclined to the fixed straight line at an angle cot -1 3. Show that 
the locus of P is a parabola. [London.] 

(74) Solve the equation 2(p- l)y = p 2 x ; show that the " jo-dis- 
criminant " is a solution of the equation, and is the envelope of the 
family of curves given by the general solution. [London.] 

(75) Obtain the differential equation of the involutes of the parabola 
y 2 = kax, and integrate it. What is the nature of the singular solution ? 

[London.] 



MISCELLANEOUS EXAMPLES 203 

(76) Prove that if the normals to a surface all meet a fixed straight 
line, the surface must be one of revolution. [London.] 

(77) Integrate the partial differential equation 

px + qy = y/{x 2 + y 2 )- 
Give the geometrical interpretation of the subsidiary integrals and 
of the general integral. [London.] 

(78) Integrate the differential equation 

dz dz 

z(x + 2y)^-z(y + 2x) ^ = y 2 - x 2 . 

Find the particular solutions such that the section by any plane 
parallel to 2 = shall be (i) a circle, (ii) a rectangular hyperbola. 

[London.] 

(79) A family of curves is represented by the equations 

x 2 + y 2 + 6z 2 = a, 2x 2 + by 2 + z 2 + ixy = /3, 
where a, /3 are parameters. 

Prove that the family of curves can be cut orthogonally by a family 
of surfaces, and find the equation of this family. [London.] 

(80) Solve b(bcy + axz)p + a(acx + byz)q = ab(z 2 -c 2 ), 

and show that the solution represents any surface generated by lines 
meeting two given lines. 

(81) (i) Solve L^ + RI = E, 

where L, R, and E are constants. 

[This is the equation for the electric current / in a wire of resistance 
R and coefficient of self-induction L, under a constant voltage E.) 

(ii) Determine the value of the arbitrary constant if / = / when 
{ = 0. 

(iii) To what value does / approximate when t is large ? 

[Ohm's law for steady currents.] 

(82) Solve L^ + RI == E cos pt. 

[The symbols have the same meaning as in the last question, except 
that the voltage E cos pt is now periodic instead of being constant. 
The complementary function soon becomes negligible, i.e. the free 
oscillations of the current are damped out.] 

(83) Find the Particular Integral of 

T d 2 Q^„dQ Q 

[This gives the charge Q on one of the coatings of a Leyden jar 
when a periodic electromotive force E cos pt acts in the circuit con- 
necting the coatings. The Particular Integral gives the charge after 
the free electrical oscillations have been damped out.] 



204 DIFFERENTIAL EQUATIONS 

(84) Show that the equations 

are satisfied by the trial solution y = nix, provided that m is a root of 

the quadratic 2 + 3m _ 16 + 3m 

7 "2 + 3m' 

dx 
and x is given by 7 -= — (2 + 3m) £ = 0. 

Hence prove that two sets of solutions of the differential equations 
are . y = 4:X = iAe 2t 

and y = -3x= -3Be~\ 

so that the general solution is x = Ae 2t + Be~ l , 

y=4Ae u -3B<r t . 

(85) Use the method of the last example to solve 

7^ + 23s-8t,-0, 

[Equations of this type occur in problems on the small oscillations 
of systems with two degrees of freedom. The motion given by y = 2x 
(or by y = -5x) is said to be a Principal or Normal Mode of Vibration. 
Clearly it is such that all parts of the system are moving harmonically 
with the same period and in the same phase. If y -2x and y + 5x are 
taken as new variables instead of x and y, they are called Principal or 
Normal Coordinates.] 

(86) Given that L, M, N, R, S are positive numbers, such that LN 
is greater than M 2 , prove that x and y, defined by 

T dx , r dy _ 
at at 

diminish indefinitely as t increases. 

[Show that x = Ae at + Be bt and y= Ee al + Fe ht , where a and b are 
real and negative. These equations give the free oscillations of two 
mutually influencing electric circuits. L and N are coefficients of 
self-induction, M of mutual induction, and R and S are resistances.] 

(87) Show (without working out the solutions in full) that the 
Particular Integrals of the simultaneous equations 



T dx ,. dy _ Cxdt T , . 
L ..-- + M / + Rx+\ — = E sin pt, 
at at J c 



,_ dx „ T dy 

dt dt J 



MISCELLANEOUS EXAMPLES 205 

are unaltered if in the first equation the term I dt is omitted and L 
is replaced by L ^. 

[This follows at once from the fact that the Particular Integrals are 
of the form A sin (pt - a). 

These equations give the currents in two mutually influencing 
circuits when the primary, which contains a condenser of capacity c, 
is acted upon by an alternating electromotive force. This example 
shows that the effect of the condenser can be compensated for by in- 
creasing the self-induction.] 

< 88 > H *£+* g4f.*-/w 

and Mp + Nf = 0, 

at at 

where LN - M z is a very small positive quantity, show that the Com- 
plementary Function for x represents a very rapid oscillation. 

[These equations occur in Rayleigh's theory of the oscillatory dis- 
charge of a condenser in the primary circuit of an induction coil with 
a closed secondary. Notice that the second equation shows that the 
secondary current is at its maximum when the primary current is at its 
minimum. See Gray's Magnetism and Electricity, Arts. 489 and 490.] 

(89) Prove that the Particular Integrals of the simultaneous equations 

m ^ = -a(x-X) + k cos ft, 

d 2 X 
M ,¥ = - AX + a(x- X) 
dr 

Bk 

may be written x = -s — 5-^ cos pt, 

a* - bB 

v -ak 

X = a?~bB C0Spt > 

where b = mj> 2 -a and B = Mp 2 - (a + A ). 

Hence show, that x and X are both infinite for two special values 
of p. 

[These equations give the oscillations of the " elastic double pen- 
dulum." Masses m and M are arranged so that they can only move 
in the same horizontal line. A spring connects M to a fixed point of 
this line and another spring connects m to M. A periodic force acts 
upon m, and the solution shows that both masses execute forced vibra- 
tions whose amplitude becomes very large for two special values of p. 
Of course this is the phenomenon of Resonance again. It is important 
to notice that the values of p that give resonance in this case are not 
the same as they would be if only one mass were present. This may 
be applied to the discussion of the " whirling " in a turbine shaft. 
See Stodola's Steam Turbine.] 



206 DIFFERENTIAL EQUATIONS 

(90) Show that the solution of the simultaneous equations 
($m + M)4a™+2Mb < ^£= T g{m + 2M)e, 

where m — M and a = b, may be expressed by saying that 6 and <p are 
each composed of two simple harmonic oscillations of periods 2ir\p x and 
27r/p 2 , Pi 2 and p£ being the roots of the quadratic in p 2 , - 

28a V - 8iagp 2 + 27 g 2 = 0. 
[These equations give the inclinations to the vertical of two rods 
of masses m and M and lengths 2a and 26 respectively when they are 
swinging in a vertical plane as a double pendulum, the first being freely 
suspended from a fixed point and the second from the bottom of the 
first. The two oscillations referred to are known as the Principal (or 
Normal) Oscillations. Similar equations occur in many problems on 
small oscillations. A detailed discussion of these is given in Routh's 
Advanced Rigid Dynamics, with special reference to the case when the 
equation in p has equal roots.] 

< 9i > % + 4 + **-°< 

d 2 y dx 
d¥~ K dt 

[These equations give the motion of the bob of a gyrostatic pen- 
dulum which does not swing far from the vertical. Notice that if the 
initial conditions are such that 5 = 0, we get motion in a circle with 
angular velocity p, while if A = 0, we get motion in a circle with angular 
velocity q in the opposite sense. (For p, q, A, B see the answers.) 

Similar equations hold for the path of revolving ions in the ex- 
planation of the Zeemann Effect (the trebling of a line in a spectrum 
by a magnetic field). See Gray's Magnetism and Electricity, Arts. 
565-569.] 

(92) Given (dx 

dT + aX = °> 

dz 7 

dt =hj > 

x + y + z = c, 

where a, b, c are constants, obtain a differential equation for z. 

dz 
Hence prove that if 2= =0 when t — 0, 

z = c + ,: \be~ at - ae~ he \. 

a-b 

[These equations occur in Physical Chemistry when a substance A 
forms an intermediate substance B, which then changes into a third 



— + c 2 !/ = 0. 



MISCELLANEOUS EXAMPLES 207 

substance C. x, y, z are the " concentrations " of A, B, C respectively 
at any time t. See Harcourt and Esson, Phil. Trans. 1866 and 1867.] 

(93) The effect on a simple dynamical system with one degree of 
freedom of any other dynamical system to which it is linked can be 
represented by the equation 

x + 1/j.x + n 2 x — X. 

If the exciting system of waves is maintained steady so that 

X = A cos pt, find the value of p for which there is resonance, and prove 

that if fx exceeds a certain value there is no resonance. Draw curves 

illustrating both cases. [Math. Trip.] 

(94) Solve the differential equation 

x + 2hx + n 2 x == when k 2 < n 2 . 
In the case of a pendulum making small oscillations, the time of a 
complete oscillation being 2 sees, and the angular retardation due to 
the air being taken as -04 x (angular velocity of pendulum), show that 
an amplitude of 1° will in 10 complete oscillations be reduced to about 
40'. [Take logi e= -4343.] [Math. Trip.] 

(95) The motion of a system depends practically on a single co- 
ordinate x ; its energy at any instant is expressed by the formula 
\mx 2 + ^ex 2 ; and the time-rate of frictional damping of its energy is 
\kx 2 . Prove that the period (t ) of its free oscillation is 

2 -( £ -r^)" 4 - 

\m 16 »iv 

Prove that the forced oscillation sustained by a disturbing force of 

e k 2 
type A cos pt is at its greatest when p 2 — ■ — ^ , and that the amplitude 

of this oscillation is then — ^°, while its phase lags behind that of the 

7T/C 

force by the amount tan -1 -^. [Math. Trip.] 

1 /ds\ 2 

(96) Show that the substitution T = - ( -j j reduces 

d 2 s D /M 2 n 

cti 2+P \dt)= Q 

dT 
to the linear form - T -+2PT = Q. 

as 

From (, + o) g + (*)". (,_«,)* 

ds 
with the conditions . =0 and s = 2a when t = 0, obtain 
dt 

dt) = lp- 2a) 
. d 2 s g 

and ^ = 3- 



208 DIFFERENTIAL EQUATIONS 

[This gives the solution of the dynamical problem : " A uniform 
chain is coiled up on a horizontal plane and one end passes over a' 
smooth light pulley at a height a above the plane ; initially a length I 
2a hangs freely on the other side. Prove that the motion is uniformly 
accelerated." See Loney's Dynamics of a Particle and of Rigid Bodies, 
p. 131.] 

(97) Find a solution of the equation 



1/^+—- (sin 0*^-0 

drV dr) + smede\ ^de)~ v 



of the form <j> =f{r) cos d, 

given that — ~- = V cos Q when r = a 

or 

and — -^- = when r — co . 

or 

\(f> is the velocity-potential when a sphere of radius a moves with 
velocity V in a straight line through a liquid at rest at infinity. See 
Ramsey's Hydro- Mechanics, Part II. p. 152.] 

(98) Find a solution of Ji=c 2 rr{ 

which shall vanish when x — 0, and reduce to A cos (pt + a) when x = b. 

[This gives the form of one portion of a stretched string, fixed at 
both ends, of which a given point is made to move with the periodic 
displacement A cos (pt + a). The portion considered is that between the 
given point and one of the ends. See Ramsey's H ydro- Mechanics, 
Part II. p. 312.] 

(99) Obtain the solution of 



~3t 2 \di* r dr) 



in the form r<p=f(ct-r) + F(ct + r). 

[0 is the velocity-potential of a spherical source of sound in air. 
See Ramsey, p. 345.] 

(100) Obtain a solution of 

dx 2 dy 2 ' 
such that d<f)/dy = when y= -h 

and (j> varies as cos (mx-nt) when y = 0. 

[0 is the velocity-potential of waves in a canal of depth h. the sides 
being vertical. See Ramsey, p. 265.] 

(101) Obtain the solution of the simultaneous differential equations 

d 2 x . di) _ 
(fr! -2„ rf/+r , = 0, 

d 2 )/ n d.r „ 



MISCELLANEOUS EXAMPLES 209 

with the initial conditions 

dx . dy . 

"-* »-<>■ s-* l=°- 

in the form z = — {(</ + w) e'^ -">' + (£ -n)e-*fa+ n )'}, 

where z = x + iy and g = \/(p a + w 2 ). 

Show that the solution represents a hypocycloid contained between 
two concentric circles of radii a and an/q. 

[This example gives the theory of Foucault's pendulum experiment 
demonstrating the rotation of the earth. See Bromwich, Proc. London 
Math. Soc. 1914.] 

(102) Obtain an approximate solution of Einstein's equation of 
planetary motion g2 u ^ 

in the following manner : 

(a) Neglect the small term 3mu 2 , and hence obtain 

u = j-^{l+e cos (0-d)}, as in Newtonian dynamics. 

(b) Substitute this value of u in the small term 3mu 2 , and hence 
obtain 

d 2 u m 3m 3 6m 3 . . 3/n 3 e 2 , n . 

dlb* +u== h 2 + ~W + l^ e cos ^~ ct) + ~W^ os 2( ^~ CT)} - 

(c) Neglect all the terms on the right-hand side of this differe itial 

equation except ,- 2 and -jj- e cos (<p - ct). The term in cos (<p - ~) mist 

be retained ; it is of the same period as the complementary function, and 
therefore produces a continually increasing particular integral. [See the 
resonance problem Ex. 36 on p. 46.] Hence obtain 

m (, . . 3m 2 . . .1 

w = /2 jl -fecos {(ji - w) + -p- e0 sin (0 -ts) Y 

AM 

= 2 {1 + e cos ((f> - & - e)} approximately, 

where e = ^- 2 - <f> and e 2 is neglected. 

[This result proves that when the planet moves through one revolu- 
tion the perihelion (given by <p - T3 - e = 0) advances a fraction of a 

revolution eiven by - = 7 o • When numerical values are given to the 
& J <p h 2 

constants it is found that Einstein's theory removes a well-known 

discrepancy between observed and calculated results on the motion 

of the perihelion of Mercury. See Eddington, Report on the Relativity 

Theory of Gravitation, pp. 48-52.] 



210 DIFFERENTIAL EQUATIONS 

(103) L(x, y, x', y') is a function of the variables x, y, x', y' . 
X, Y are denned by the equations 

X-— y=— 

dx" dy' 

If these equations can be solved for x' and y' as functions of X, Y, x, y, 
and if H (X, Y, x, y) is the function obtained by expressing 

Xx' + Yy'-L 

entirely in terms of X, Y, x, y, then prove that 

£-* 4 

and IST'to (2) 

Prove also that the equation 

dAdx'J dx K} 

is transformed into -=- = - -=— (4) 

at ox 

[This is the Hamiltonian transformation in dynamics. Equation (3) 
is a typical Lagrangian equation of motion in generalised co-ordinates. 
Hamilton replaces it by the pair of equations (1) and (4). See Routh's 
Elementary Rigid Dynamics, Chap. VIII. This transformation should 
be compared with that of Ex. 21 of the miscellaneous set at the end of 
Chap. XII., where we had two partial differential equations derivable 
from each other by the Principle of Duality.] 

(104) Show that Jacobi's method (Art. 140) applied to Hamilton's 
partial differential equation 

^ + H(x v x 2> ... x„, p v p 2 , ... p„, «) = 

dx,. BH dp r dH 

leadSt ° W = Wr' ^ = 'dx r < r==1 > 2 '- W >' 

which are the equations of motion of a dynamical system, in Hamilton's 
form. [See Whittaker's Analytical Dynamics, 2nd ed., Art. 142.] 

(105) (i) Prove that if u(x, y, z) = a 
and v(x, y, z) = b 

are any two integrals of the system of differential equations 
dx dy dz 



p(x,y,z) q(x,y,z) r(x,y,z)' 
v) 1 d(u, v) 1 d(u, v) 
z) q d(z, x) r d{x,y) 
[m is called a multiplier of the system.] 



1 dlu, v) 1 d(u, v) 1 d(u, v) , . 

then - -7 r = - a7 r = - , T — .- = m (x, y, z), say. 

p o{y, z) q d(z, x) r d(x, y) 



MISCELLANEOUS EXAMPLES 211 

(ii) Show that m satisfies the partial differential equation 

(iii) If n(x, y, z) is any other multiplier of the system, show that 

d /m\ d /m\ d /ra\ _ 

dx\n/ dy\nJ dz\nJ ' 

u ( IYI iVh It 1)) 

and hence that -W— 2 — '-r- = identically, 

d(x,y,z) 

so that m/n is a function of u and v, and m/n = c is an integral of the 

original system of differential equations. 

(iv) If u(x, y, z)=a can be solved for z, giving z=f{x, y, a), and 

if capital letters V, P, Q, R, M denote the functions of x, y, a, obtained 

by substituting this value of z in v, p, q, r, m, then prove that 

doc du 
V(x, y, a) = 6 is an integral of — = ~ . 

BV du 
Prove also that MP = — -=— «- 

ay oz 

7i^ dV du 

and MQ= Txdz 

where =- is to be expressed in terms of x, y, a), so that 

dV = M(Qdx-Pdy)l d d ". 

[This suggests that if any integral u=a and any multiplier m are 

known, then M(Qdx-Pdy)l ^ will be a perfect differential, leading 

to an integral of the system when a is replaced by u(x, y, z). 

For a proof of this theorem see Whittaker's Analytical Dynamics, 
2nd ed., Art. 119. A more general theorem is that if (n-1) integrals 
of a system of differential equations 

dx x dx 2 _ _ dx n _ dx 

V\~ V2~'"~ Pn V 
are known and also any multiplier, then another integral can be deter- 
mined. This is generally referred to as the theorem of Jacobis Last 
Multiplier. In Dynamics, where this theorem is of some importance 
(see Whittaker, Chap. X.), the last multiplier is unity.] 
(v) Show that unity is a multiplier of 

dx dy dz 

xz -2y 2x- yz y 2 - x 2 
and s 2 +?/ 2 + 2 2 = aan integral, say u(x, y, z)=a. 
Show that in this case 

M(Qdx-Pdy)l^ = d{-lxy-V(a-x 2 -y z % 

and hence obtain the second integral xy + 2z = b. 



212 DIFFERENTIAL EQUATIONS 

n 

(106) Show that if y = 1 e*'/(0 dt, where a and 6 are constants, then 

J a 

• x ^{^ y+ ^{T^ y=ehx ^ {h)f{h) ~^ x ^ {a)f{a) 

-(W(O/'(«)+0'(O/W-^W/«}& 

J a 

Hence prove that y will satisfy the differential equation 

if ^>W/(0 = exp{J^|^} 

and e bx ( p(b)f(b)=0 = e tlx (f>(a)f(a). 

Use this method to obtain 

^ J -oo V(^-l) J-l V(' 2 -l) 

as a solution, valid when x>0, of 

The corresponding solution for the case x<0 is obtained by taking 
the limits of the first integral as 1 to oc , instead of -co to -1. 

[Exs. 106-108 give some of the most important methods of obtaining 
solutions of differential equations in the form of definite integrals.] 

(107) Verify that v = v + ^-\ e~*dz 

, ,. , dv d 2 v 

is a solution of ~= K ~-, 

ot ox 1 

reducing, when t = 0, to v + V for all positive values of x and to v - V 
for all negative values. 

[v is the temperature at time t of a point at a distance x from a 
certain plane of a solid extending to infinity in all directions, on the 
supposition that initially the temperature had the two different constant 
values v + V and v - V on the two sides of the plane x = 0. 

Kelvin used this expression for v in his estimate of the age of the 
earth (see Appendix D of Thomson and Tait's Natural Philosophy). The 
discovery that heat is continually generated by the radio-active dis- 
integration of the rocks introduces a new complexity into the problem.] 

(108) (a) Show that 



V= [ e lx+m y +nz f(s, t) ds dt 



(the limits being any arbitrary quantities independent of x, y. z) is a 
solution of the linear partial differential equation with constant 
coefficients / a a 7)\ 

F[ , , J V=0 

\dx dy dz/ 



MISCELLANEOUS EXAMPLES 213 

if I, m, n are any constants or functions of s and t such that 

F(l, m, n) = 0. 
Extend the theorem to the case when there are n independent 
variables x, y, z, ... , and (n - 1) parameters s, t, ... . 

Obtain V= f #l xcoat+ ** int + u )f(8, t) ds dt 

as a solution of ^ + -~~2 = a - ' P 1 - Todd.] 

(P) r) F)\ 
«-,_-, p-j F is a homogeneous linear 

partial differential equation with constant coefficients a solution is 



]/d 



x + my + nz, t) dt, 



where the limits are any arbitrary quantities independent of x, y, z, and 
I, m, n are any constants or functions of t such that 

F(l, m, n) = 0. 
Extend the theorem to the case when there are n independent 
variables and (n - 2) parameters. [See H. Todd, Messenger of Mathe- 
matics, 1914.] 

n. 
Obtain 7=1 f(x cos t + y smt + iz, t) dt 

d 2 V d*V d 2 V . 
as a solution of = .,- + -=-=• + -=-r- =0. 

ox* oy £ oz l 

[Whittaker's solution of Laplace's equation.] 

(109) By substituting the trial solution 

a, a 9 
y = a o + x + x% + ~' 

.._ . , . dy 1 

in the differential equation -=- + y = - , 

0! 1! 2! 3! 
obtain the series y = — I- „ + + .+.... 

x x £ x 6 xr 

Prove that this series is divergent for all values of x. 

Obtain the particular integral 

y = e~ x I — dx, 

J -00 X 

and by repeated integration by parts show that 

C' e* 7 0! 1! 2! n! , f ' (n+l)!e* 

e x \ —dx = - + - 9 +-= + ...+ -— i + e J — -is dx. 

j_ M X X X? X A X n+1 J -r. Z»+ 2 

Hence prove that if x is negative the error obtained by taking n 
terms of the series instead of the particular integral is less than the 
numerical value of the (n + l) th term. 

[Such a series is called asymptotic. See Bromwich's Infinite Series, 
Arts. 130-139.] 



214 DIFFERENTIAL EQUATIONS 

(110) Show that if the sequence of functions f n (x) be defined by 

f (x)=a + b(x-c), where a, b, c are constants, 

and /„(*) = JV*)^(0/«-i(0^ 

d 2 
then dx^ n{x) = ~ F{x) f»-i( x )' 

00 

Hence show that y = ^f n {%) is a solution of 
o 

provided that certain operations with infinite series are legitimate (for 
a proof of which see Whittaker and Watson's Modern Analysis, p. 189. 
They give a proof of the existence theorem for linear differential equa- 
tions of the second order by this method). 

(111) Prove that the solution of the two simultaneous linear differ- 
ential equations with constant coefficients 

f(D)x + F(D)y=0, 

<t>(D)x + \f,(D)y=0 

(where D stands for d/dt), may be written 

x = F(D)V, 
y=-f(D)V, 

where V is the complete primitive of 

{f(D)yl,(D)-F(D)<t>(D)}V=0. 

Hence show that if the degrees of/, F, 0, \js in D be p, q, r, s respec- 
tively, the number of arbitrary constants occurring in the solution will 
in general be the greater of the numbers (p + s) and (q + r), but if 
(p + s) — (q + r) the number of arbitrary constants may be smaller, and 
may even be zero as in the equations 

(D + l)x + Dy=0, 

(D + 3)x + (D + 2)y = 0. 

(112) (a) Prove that if y = u(x), 

y = v(x) 
are any two solutions of the linear differential equation of the first order 

P(x) yi +Q(x)y=0, 
then (t'u x - uvj/u 2 = 0, 

so that v = au, where a is a constant. 
(b) Prove that if y = u(x), 

y = v(x), 
y = w(x) 



MISCELLANEOUS EXAMPLES 215 

are any three solutions of the linear differential equation of the second 
order P{x)y 2 + Q(x) yi + R{x)y = 0, 

then P -=- («Wi - vw x ) +Q(wv 1 - vwj) = 

and P -j-iuvi-vu^+Qi^-viij) = 0. 

Hence show that w = au + bv. 

[By proceeding step by step in this manner we may show that a 
differential equation of similar form but of the n th order cannot have 
more than n linearly independent integrals.] 

(113) Let u, v, w be any three functions of x. 

Prove that if constants a, b, c can be found so that y=au + bv + cw 
vanishes identically, then 

U V w 

u x v x w x =0, 

u 2 v 2 w t 

while conversely, if this determinant (the Wronskian) vanishes, the 
functions are not linearly independent. 

Extend these results to the case of n functions. 

[Consider the differential equation of the second order formed by 
replacing u, u v u 2 in the determinant by y, y v y 2 respectively. Such 
an equation cannot have more than two linearly independent integrals. 

The Wronskian is named after Hoene Wronski, one of the early 
writers on determinants.] 

(114) Prove that z = e ix ( t ~ 1/t ) satisfies the partial differential equation 
Hence, if J n {x) is defined as the coefficient of t n in the expansion 

e ^-m^-^t n J n (x), 

prove that y = J n (x) satisfies Bessel's equation of order n, 

[The operations with infinite series require some consideration.] 

(115) If u j: denotes a function of x, and E the operator which changes 
u x into u x+1 , prove the following results : 

(i) Ea x = a . a x , i.e. {E-a)a x = Q. 

(ii) E 2 a x = a 2 . a x . 

(iii) E(xa x ) = a{xa x )+a . a x , i.e. (E-a)(xa x )=a . a*. 

(iv) (E-a)*(xa x )=0. 

(v) (p E 2 +p 1 E+p 2 )a x = (p Q a 2 +p 1 a+p 2 )a x , if thep's are constant. 



216 DIFFERENTIAL EQUATIONS 

(vi) u x =Aa x +Bb x is a solution of the linear difference equation 
PoU x+z +PiU x+1 +p 2 u x = 0, 
i.e. {p E 2 + p 1 E+p 2 )u x =0, 
if A and B are arbitrary constants and a and b the roots of the auxiliary 
equation p m 2 + ^m^- p 2 = 0. (Cf. Art. 25.) 
Solve by this method (2£ 2 + 5E + 2) u x = 0. 

(vii) u x = {A + Bx)a x is a solution of (E 2 -2aE + a 2 )u x = 0. 
Here the auxiliary equation m 2 -2am + a 2 = has equal roots. 
(Cf. Art. 34.) 

(viii) u x = r x (P cos xd + Q sin x6) is a solution of 

(Po E2 +Pi E + Pz) u x = ^ 
if P and Q are arbitrary constants, p±iq the roots of the auxiliary 
equation p m? +p 1 m+p 2 = 

and p + iq = r (cos + i sin 6). (Cf. Art. 26.) 

Solve by this method (E 2 -2E + i)u x = 0. 

(ix) The general solution of a linear difference equation with constant 
coefficients 

F(E)=(p E n +p 1 E"-i + ...+p n _ 1 E + p n )u x =f(x) 

is the sum of a Particular Integral and the Complementary Function, 
the latter being the solution of the equation obtained by substituting 
zero for the function of x occurring on the right-hand side. (Cf. 
Art. 29.) 

(x) a x /F(a) is a particular integral of 
F(E)u x = a x , 
provided that F(a)=f=0. (Cf. Art. 35.) 

Solve by this method (E 2 + 8E - 9) u x = 2 x . 

[For further analogies between difference equations and differential 
equations, see Boole's Finite Differences, Chap. XL] 



^ 



ANSWERS TO THE EXAMPLES 

CHAPTER I. 
Art. 5. 

(5) The tangent to a circle is perpendicular to the line joining the 

point of contact to the centre. 

(6) The tangent at any point is the straight line itself. 

(7) The curvature is zero. 

Art. 8. 

Sf*2 /y3 /y>4 

(1) y = a + ax + a^ + a^ + a-rj + ...=ae x . 

(2) y = a + bx - a —^ - 6 — } + a jy + . . . = a cos x + b sin x. 

Miscellaneous Examples on Chapter I. 

< 3 > 3-« *+*-* 

(^>»4IW{-(in]-V{-(2)T «3-* 
« Mg)T=°<g) a . - >*- • 

(12) y = ae x + ber*. (14) 60° and -60°. 



V 



ii DIFFERENTIAL EQUATIONS 

(15) Differentiate and put x = l, y = 2. This gives ~ and hence p. 
(17) (i) z + l=0; (ii) y 2 = x 2 + 6x + l. 

CHAPTER II. 
Art. 14. 

(1) 6x 2 + 5xy + y 2 -9x - Ay = c. (2) sina;tan?/ + sin(a; + ^) = c. 

(3) sec x tan y - e x = c. (4) x-y + c = \og(x + y). 

(5) x + ye x3 = cy. (6) y = cx. 

(7) e ?/ (sina; + cosa;)=c. (8) x*y + icy + 4 = 0. 

(9) ye x -cx. (10) sin a; cos y = c. 

Art. 17. 

(1) (x + y)z = c{x-y). ^$2) z 2 + 2?/ 2 (c + log y)=0. 

*(3) a;?/ 2 = c(a;-?/) 2 . (4) ca; 2 = ?/ + -v/(cc 2 + ?/ 2 ). 

(5) (2x-y) 2 = c{x + 2y-5). (6) (z + 5?/-4) 3 (3a: + 2?/ + l) = c. 
(7) x-y + c = log(3x-4:y + l). (8) 3z-3?/ + c = 2 log(3x + 6y-l). 

Art. 21. 

(1) 2y = (x + a) 5 + 2c(x + a) z . (2) x# = sin x + c cos x. 

(3) 2/ log x =7Jtag\)^£. (4) cc 3 = ^/ 3 (3sina;4-c). 

! (5) y 2 (x + ce x ) = l. (o) x = y 3 + cy. (7) a? = e~ ! '(c + tan y). 

Art. 22. 

(1) The parabola y 2 = 4ax + c. 

(2) The rectangular hyperbola xy = c 2 . 

(3) The lemniscate of Bernoulli r 2 = a 2 sm20. 

(4) The catenary y = k cosh -^— . (5) xy — c 2 .. \J 

(6) i/ S = a; § + c § . (7) yv = c&. (8) r 2 = ce^. 

(9) log r + 10 2 + JO 3 - c. (10) The equiangular spirals r = ce ±e tan a . 

Miscellaneous Examples on Chapter II. 

V (1) xy = y* + c. (2) cx* = y + ^/(y 2 -x 2 ). 

(3) sin xsiny + e 8in r = c. (4) 2x 2 - 2xy + 3y + 2cx 2 y = 0. 

(5) cxy = y + \/(y 2 -x 2 ). (11) x 3 y~ 2 + 2x 5 y~' i = c. 
(12) tan- 1 (^)+log(«/2/)=c. (14) (x 2 -l+y*)e x * = c. 

(15) (i) The Reciprocal Spiral r(0- a ) = c. 
(ii) The Spiral of Archimedes r = c(6-a). 

(16) The parabola Sky 2 = 2x. (18) x = y(c -k log y). 



ANSWERS 



in 



(19) (i) a? + (y-c) 2 = l +C 2 , a system of coaxal circles cutting the given 
system orthogonally, 
(ii) r 2 = ce- 63 . (id) n 2 = r{c + log(cosec nQ + cot n$)}. 



<*» ("'gX-'D-f -»• 



(21) log(2a; 2 ±an/ + */ 2 )+7^tf 



_!»±2y 
xy/1 



=o. 



4 



CHAPTER III. 



Art. 28. 

(2) y = A cos 2x + B sin 2x. 
(4) y = e 2x ( J. cos a; + B sin a;) 
(6) s = A + Be- 4t . 
(8) ?/ = 2e- a! -e- 2a! . 



(1) y^^e-^ + ^e" 3 *. 

(3) y = Ae~ 3x + Be-* x . 

(5) s = e- 2t (^cos3« + 5sin30. 

(7) y = Ae x + Be- x + Ce~ 2x . 

(9) ?/ = 4 cos (2a; -a) + 5 cos (3a; -/3). 

(10) y = A cosh (2a? - a) + £ cosh (3x - /3), or - 
y - Ee 2x + Fe~ 2x + Ge* x + He~ 3x . 

(11) y = Aer** + Be* cob {xy/3- a). 

(12) y = Ae 2 * + Be" 2 * + Ee~ x cos (aj-y/3 - a) + Fe x cos (x V3 - /3) 

(13) 6 = a cos *vW- ( 14 ) ^ 2 < 4mc - 



(16) Q-Qf-nWcoBt^^- 



R 2 \ 
iL 2 )' 



Art. 29. 

(1) y = e x (l+Acosx + Bsinx). (2) ?/ = 3 + 4e a; + Be 12 *. 
(3) y = 2 sin 3a; + A cos 2a; + B sin 2a;. (4) a = 2; 6 = 1. 

(5)a-6;ft--l. (6) a = /£; p = 2. (7) a = l .; 6 = 2; y-1. 

(8) a = 2. (9) 4e 3 *. (10) 3e 7 *. 

(11) -f sin 5a;. (12) £ cos 5a; - £ sin 5x. (13)2. 

Art. 34. 

(1) y = A + Bx + (E+Fx)er*. 

(2) ?/ = (A + Bx + Cx 2 ) cos x + (E + Fx + Gx 2 ) sin a;. 

(3) «/ = (.4 + Z?a;) e* + 2? cos a; + .F sin a;. 

(4) y = A + Bx + Ce x + {E + Fx)e-^.. 

Art. 35. 

(1) y = 2e Zx + e~ 3x (A cos 4a; + Bsin 4a;). 

(2) y = e--P*(4 cos qx + B sin </x) + e ax /{(a + p) 2 + q 2 }. 

(3) y = (A+9x)e 3x + Be- 3x . 

(4) «/ = 4+(B + |j)e a: + (C + |a;)e- a: . 



IV DIFFERENTIAL EQUATIONS 

(5) y - [A + ax/2p) cosh px + B sinh px. 

(6) y = A+(B + Cx-2x 2 )e- 2x . 

Art. 36. 

(1) y = 2sin 2a;-4cos2x + ile- a! . 

(2) */ = 4 cos 4a; - 2 sin ix + Ae 2x + Be 3x . 

(3) y = 2co&x + e-* x {Aco&Zx + BsmZx). 

(4) !/ = sin 20a; + e- a! (^ cos 20a; + 5 sin 20a;). 

Art. 37. 

(1) y = x 3 -3x 2 + 6x-6 + Ae~ x . (2) y = 6x 2 -6x + A + Be~ 2r . 

(3) y^Gx + e + ^ + ^e 3 ^ 

(4) </ = a; 3 + 3a; 2 + ^a; + i? , + (^ + J Ba;)e 3a; . 

(5) y = 24x* + Ux-5 + Ae- x + Be 2x . 

(6) # = 8a; 3 + 7a; 2 - 5ic + .4er* + Be 2x + C. 

Art. 38. 

(1) 2/ = ^4 cosa; + (l? + 2a;)sina;. (2) y = Ae x + (x + 2)e* x . 

(3) y = ^4e 2a: + ( B + Gx - 20a; 2 - 20a; 3 - 15a; 4 - 9a^) e"*. 

(4) y = {Asinx + (B-x) cosx}e~ x . 

(5) y = (^ + Bx - x 3 ) cos x + (E + Fx + 3x 2 ) sin x. 

(6) s/ = ^ + (£ + 3a;)e a! + Ce- K + a; 2 + £cosa; + (jF + 2a;)sina;. 

(7) y ={A sin 4a; + (B- x + x 2 ) cos 4a;} e 3 *. 

Art. 39. 

(1) ?/ = ^a; + J5x 2 + 2x 3 . 

(2) y = 2 + ^lar 4 cos (3 log x) + Bx~* sin (3 log x). 

(3) y = 8 cos (log a;) - sin (log as) + via;" 2 + Sx cos (^/3 log a; - a). 

(4) i/ = 4 + log x + Ax + Bx log x + Cx (log x) 2 + D'x (log a;) 3 . 

(5) y = (l + 2a;) 2 [{log (1 +2x)} 2 + ^ log (1 +2a;) + B]. 

(6) y = A cos {log (1 + a;)-ct} + 2 log (1+a;) sin log (1+x). 

Art. 40. 

(1) y^A cos (x-a) ; z—-Asin(x-a). 

(2) y = Ae^ + Be 3x ; z = 6Ae 5x -7Be 3x . 

(3) y = Ae x -?B cos (2x -a); z = 2Ae x -B cos (2x - a). 

(4) ?/ = e a; + ^ + J Be- 2a; ; 2 = e x + A - Be~ 2x . 

(5) y = A cos (x-a) + 45 cos (2a; - ft) + cos 7a; ; 
z = A cos (x - a) + B cos (2a; - (3) - 2 cos 7a;. 

(6) y = - S^e 3 * - 4 J Be 4 * + 2e-* + cos 2a; - sin 2a; ; 
z = Ae 3x + Be* x + 3e~ x + 4 cos 2a; + 5 sin 2a;. 



ANSWERS 

Miscellaneous Examples on Chapter III. 

y=>{A + Bx + Cx 2 )e x + 2e? x . (2) y = {A + Bx + 6x*)e-**fi. 

y = Ae~ Zx + Be~ 2 x + Cer x +Ex + 2e- 2 *(sin x - 2 cos a;). 

y 4Ae x + B cos (2x - a) - 2e ie (4 sin 2x + cos 2a?). 

y = (A + Bx + Gx i )e- x + (E + x + 2x 2 )e 3x . 

y= A &in(x -a) + B sinh (3x — yS) — 2 sinh 2x. 

y = (A + Bx + 5x 2 ) cosh x + (E + Fx) sinh x. 

y m 3 + 4x + 2X 2 + (A + Bx + 4x 2 ) e 2x - cos 2x. 

?/ = ( J. + Bx + 3 sin 2a; - 4x cos 2a; - 2a; 2 sin 2a;) e 2x . 

y = A cos (x - a) + f - ^- cos 2a; - J x cos a; + T V sin 3a;. 

y = A cos (a; - a) + B cos (3a; - f3) - 3aTcos x + x cos 3a;. 

y = (A + A x x + A $? + . . . + Aa^x*' 1 ) e ax + a x /(\og a - a) a . 

y = A + B\ogx + 2(\ogx) 3 . (14) y = A + B x^ + jx 2 . 

y = Ax~ 3 + B cos (\/2 log x - a). 

y = A + B log (x + 1) + {log (x + 1)} 2 + x 2 + 8x. 

x = A£ l + Be~ zt + E cos t + F sin t - e l ; 

y = Ae st + 25e~ 3t + (32? - 42?) cos « + (3F + 42?) sin « - e ( . 

x = ^4e 2 < + Be~ l cos ( V3J - a) ; 

y = Ae 21 + Be~ l cos ( ^/3t - a + 2tt/3) ; 

e = Ae 21 + Be- 1 cos ( V& - a + 4tt/3). 

x = ^ + 2?H; y = Bt- x -At. 

x = At cos (log £ - a) + 2ft -1 cos (log t-fi); 

,y = At sin (log £ - a) - 2ft -1 sin (log £ - /S). 

(i) (x-l)e 2a; ; (ii) £(x 2 -2a; + l) sin x + \{x 2 - 1) cos x. 

y = e 2x + Ae x . 

y = (sin ax)/(p 2 -a 2 )+A cos px + B sin yx. 

?/ = .4e aa: + J8e 6a: + e 6x xe- 6x (log x - 1) dx. 

(iii) ?/ = J. cos (x - a) - x cos x + sin x log sin a;. 

(i) k/(2phe) ; (ii) zero. 

2/ = E cos nx-vF sin «x + (? cosh wx + H sinh wx. 

CHAPTER IV. 
Art. 42. 

a* 82: 

(1) dy = a dx' 

d 2 z d 2 z 

(2) sr-s + =-s = 0. (Laplace's equation in two dimensions.) 
ox 2 ay 2 

.,, a 2 * a 2 * 1 a 2 z ... a? a? 

(3) ax 2+ av 2= o 2 a^ W»s +ap *-°" 



VI DIFFERENTIAL EQUATIONS 

(5) b^- + a=- = 2abz. „ . 

ox dy \ 

^ X dx +y dv = nZ ' ( Euler ' s theorem on Homogeneous Functions.) 

Art. 43. 



(1) 


3*z 

dx 2 


dz 

-w < 2 > 


d 2 z d 2 z d 2 z 
dx 2 + df + W 


= 0. 


(3) 


dz dz 
dx dy 


-1 


(4) 


z = 


■dz dz (dz\< 
V dx + y dy + \dx) 


<&■ 














(9) 


4:Z = 


(dz\ 2 /dz\ 2 
\dx) + \dy)' 




(6) 


dz dz 
dx dy 











Art. 45. 

(1) y = Ae-P(*+ t ). (2) z = A sin px sin pay. (3) z = A cos p(ax-y). 

(4) V = Ae-P x +w sin Zy/{p 2 + q 2 ), where p and q are positive. 

(5) V = C cos (pqx + p 2 y + q 2 z). 

(6) V = Ae~ rt sin (mw/Q sin (mry/l), where m and n are any integers 

and rl 2 = 7r 2 (m 2 + w 2 ). 

Art. 48. 

(1) — (sm x + ^sin 3a; + isin 5x + ...). 

7T 

(2) 2(sin a- a sin 2x + ^sin 3x~ ...). 

2|"/7r 3 6tt\ . /tt 3 6tt\ . /tt 3 6tt\ . n 1 

41" 2 4 6 1 

( 4 ) - \jpZi sln 2a; + pTTi sin 4x + gaTi sin 6» + . . . J. 

2 

(5) - [|(1 + e*) sin » + f (1 -e') sin 2^4-^(1 +e") sin 3z 

x H-^l-e*) sin ix + ...]. 

/*» 32 ^> * . Mr/. . Mr Mr\ . 

(6) — 2i ^3 sm 2" I 4 Sln T " W7r cos T/ sln nx " 

(7) (a) (2), (3), and (6) ; (6) (6). 

Miscellaneous Examples on Chapter IV. 

... d 2 V 1 dV d 2 V a 2 d f 2 dV\ 

(7) V = V e"J x sin (nt -gx), where g = + ^(n/2K). 

(12) F = - (e- A ' ( sin B + ^ye -5 '*' sin 3a; + y i^e- 25A7 + ...). 

7T 

(13) Replace x by ttz/Z, t by tt 2 /// 2 , and the factor 8/tt by 8P/tt 3 . 



ANSWERS vii 

(14) V = Z~ - (e~ iKt cos 2x + \e-™ Kt cos ix + ler*** 1 cos 6x + ...). 

b 

(15) V = — (e~ Kt sin x + $e-° Kt sin 3x+ le-™ Kt 8in 5x + ...). 

[Notice that although 7 = 100 for all values of x between 
and ir, V = for x = or 7r, a discontinuity.] 

(16) Write 100- V instead of V in the solution of (15). 

4F« 

(18) F = — & {e-*W cos{ttx/21) + ^r 9W /« ! cos (3?re/2Z) + ...}. 

7T 

(19) ?/ = — (sin x cos ttf - A sin 3a; cos 3?tf + ^ sin 5x cos 5vt - ...). 

7T 

(22) gj^ = 0; y=f(x-at) + F(x + at). 

CHAPTER V. 
Art. 52. 

(1) (y-2a?-c)(y + 3a:-c)=0. (2) (2?/-:r 2 -c)(2,>/ + 3:r 2 -c)=0. 

(3) 49(t/-c) 2 = 4x 7 . (4) (2?/-cc 2 -c)(2x-2/ 2 -c)=0. 

(5) (2y - x 2 - c) (y - ce x ) (y + x-1 - ce~ x ) = 0. 

(6) (y-e x -c){y + er x -c)=Q. 

Art. 54. 

(The complete primitives only are given here. It will be seen later 
that in some cases singular solutions exist.) 

(1) z = 4^ + 4^ 3 ; y — 2p 2 + 3p 4 + c. 

(2) x = ±(p + p~ 1 ); y = }p 2 -%\ogp + c. 

(3) (p-l) 2 x = c-p + logp ; (p-l) 2 y = p 2 (c-2 + \ogp)+p. 

(4) x = ±p 2 + 3^ + 3 log (p-l)+c ; y = jP + %p 2 + 3p + 3 log (p-\) + c. 

(5) x = 2 tan _1 p - p~ x + c ; y = log (p 3 + p). 

(6) a5=p + ce - ^; y = %p 2 + c (p + l) e~i>. 

$ (7)# = 2p + cp{p 2 -iy h ; y=:p 2 -l+c(p 2 -l)~ h . 
■' (8) z=smp + c; y—p sin j) + cos p. 
(9) aj = tan^» + c; y = pta,np + \ogcos p. 

(10) cc = log (p + l) -log (p-l)+log^ + c ; y = p-\og (p 2 - 1). 

(11) x = p/(l+P 2 )+tan-V, 2/ = c-l/(l+ ? ; 2 ). (12) c = l. 



CHAPTER VI. 
Art. 58. 

(1) C.P. (y + c) 2 = x z ; x = is a cusp-locus. 

(2) C.P. {y + c) 2 = x-2 ; S.S. a: = 2. 



yiu 



DIFFERENTIAL EQUATIONS 



(3) C.P. tf+cy + c 2 ^; S.S. y 2 = 4=x*. 

(4) C.P. c*(y + cosx) -2c sin x + y -cos x = 0; S.S. ?/ 2 = l. 

(5) C.P. (2x* + 3xy + c) 2 -4(x* + y)* = 0; x 2 + y = is a cusp-locus. 

(6) C.P. c 2 - 12cxy + 8cf - \2x 2 y 2 + 1 Qx 3 - ; y 2 - x = is a cusp-locus. 

(7) C.P. c 2 + 6cx«/ - 2c?/ 3 - a?(3?/ 2 - a;) 2 = ; «/ 2 + x = is a cusp-locus. 

Art. 65. 

(1) C.P. (y + c) 2 = x(x-l){x-2); S.S. x(x-l)(x-2)=0 ; z = l-l/y3 

is a tac-locus and x = l + l/\/3 a tac-locus of imaginary points 
of contact. 

(2) C.P. (y + c) 2 = x(x-l) 2 ; S.S. z = 0; a; = 1/3 is a tac-locus; x = l 

is a node-locus. 

(3) C.P. y 2 -2cx + c 2 = 0; S.S. ?/ 2 = a: 2 . 

(4) C.P. x 2 + c(x-3y)+c 2 = 0; S.S. (3y + a)(y -x)=0. 

(5) C.P. y-cx a -c 2 = 0; S.S. z 4 + 4?/ = 0; a; = is a tac-locus. 

(6) C.P. y = c(x-c) 2 ; ?/ = is a S.S. and also a particular integral; 

27*/-4a; 3 = 0isaS.S. 

(7) Diff . Eq. p 2 y 2 cos 2 a - 2pxy sin 2 a + y 2 -x 2 sin 2 a = ; 

S.S. y 2 cos 2 a = x 2 sin 2 a ; y = is a tac-locus. 

(8) Diff. Eq. (a; 2 -l)j9 2 -2a^-a; 2 = 0; S.S. x 2 + */ 2 = l; 

x = is a tac-locus. 

(9) Diff. Eq. (2x 2 + l)p 2 + (x 2 + 2xy + y 2 + 2)p + 2y 2 + l=0; 

S.S. x 2 + 6xy + y 2 = 4 ; # = ?/ is a tac-locus. 
(10) Diff. Eq. p 2 {l-x 2 )-(\-y 2 )=0; S.S. x= ±1 and y= ±1. 



oMj fab: 



Art. 67. 

S.S. z 2 + 4?/ = 0. 
(2n^ W^ 3 ; S.S. 27/ + 4a: 3 = 0. 

(3) C.P. y — cx + cos c ; S.S. (y - x sin _1 a;) 2 = 1 - x 2 . 

(4) C.P. y = cx + ^/(a 2 c 2 + b 2 ) ; S.S. x 2 \a 2 + xj 2 \h 2 = \. 

(5) C.P. y = cx-e c ; S.S. ?/ = a;(log a;-l). 

(6) C.P. ?/ = ca:-sin- 1 c; S.S. y = V(x 2 -1) -sin-V(l ~V^)- 

(7) \{y — px) 2 — —pk 2 ; 2xy = k 2 , a rectangular hyperbola with. 

axes as asymptotes. 

(8) (x - y) 2 -2k(x + y)+ k 2 — 0, a parabola touching the axes. 

(9) The four-cusped hypocycloid a; 3 + y* —k*. 




Miscellaneous Examples on Chapter VI. 

(1) No S.S. ; a; = is a tac-locus. 

(5) 2y=±3x represent envelopes, y — is both an envelope and a 
cusp-locus. 



ANSWERS IX 

(6) C.V.xy = yc + c*. 

(7) C.P. x = yc + xyc>; S.8. y=0, y + 4x*=0. (Putt/ = 1/F; 3 = 1/*.) 

(8) (i) Putting p + x = W we get 

2x = 3 (* 3 - I 2 ) ; 40y = 9 (5« 6 + 2« 5 - 5f 4 ) + c. 
(ii) C.P. «/ 2 + 4^ = 1 +2cx; S.S. a£-4y* + 4=0; ?/ = is a tac-locus. 
(11) C.P. r = a{l +cos (6 -a)}, a family of equal cardioids inscribed in 
the circle r = 2a, which is the S.S. The point r = is a cusp- 
locus. 

CHAPTER VII. 
Art. 70. 

(1) y=logsec x + ax + b. (2) x = a + y + b\og (y-b). 

(3) ay = cos (ax + b). (4) x = log {sec (ay + b) + tan (ay + b)} + c. 

(5) y = x 3 + axlog x + bx + c. 

(6) y= -e x + ae 2x + bx n - 2 + cx n ~ z + ... +hx + k. 

(7) The circle (x - a) 2 + (y - b) 2 = k 2 . The differential equation ex- 

presses that the radius of curvature is always equal to k. 

(9) \/(l + 2/i 2 ) = ky.} ; the catenary y-b = k cosh {(# - a)/k}. 

Art. 73. 

(1) y = x(a\ogx + b). (2) i/ = axcos (2 log x) +&csin (2 log a). 

(3) y = x(a\ogx + b) 2 . (4) y = x 2 (a log x + b) 2 . 

Art. 74. 

(1) y=±coth X -~^. (2) y- -log(l-x). (3) y = sm- 1 x. 

(5) (i) The conic m = yu//t 2 + (1/c - /x//t 2 ) cos B \ 
(ii) cm = cos 0\ ; (l - nlh 2 ) or cosh 6^(m/h 2 -l), according as /u>/* 8 . 

Art. 75. 

(1) y = a(x 2 + l)+be~ x . (2) y = a(x-l)+kr*. 

(3) y = a(x-l)+be~ x + x 2 . (4) y = l +e~^. (5) y-e 8 ". 

Art. 77. 

(2) y = x z + ax-l\x. (3) y = (<c 2 + ace) e* + to. 

(4) y = e 2x + (ax* + b)e x . (5) y = ax 3 + tar 3 . 

(6) «/ = ax 2 + b sin as. 

Art. 80. 

(1) y = (a - x) cos x + (6 + log sin x) sin x. 

(2) y= ja-log tan ( j +ajj J- cos 2x + &sin 2x. 




X DIFFERENTIAL EQUATIONS 

(3) y = {a-er* + \og{l+er")}e x + {b-log{l+e x )}er x . 

(4) y = ax + bx~ 1 + (l-x- 1 )e x . (5) y = ae x + (b-x)e? x + ce? x . 

Miscellaneous Examples on Chapter VII. 

(1) y = ae e ' h -b. (2) y = a + \og{x 2 + b). 

o^n+i x n x n— * 

(3) v = ; r TT + 2a— , + a 2 -. — , + bx n ~ 2 + cx n ~ z + . . . + hx + k. 

v * (n + l)\ n\ {n-\)\ 

(4) y = - 3 2 ~ n cos {3x - \tt ( n - 2)} + a cos a; + 6 sin x + ex" -3 + . . . + foe + k. 

(5) t/ = ax + b log x. (6) y = ae x + b(x 2 -l)e* x . 

(7) w = a cos nx + b sin nx + ~ sin was — ^ cos wx log sec nx. 
v ' * n n 2 

(8) ?/(2a; + 3) = aloga; + 6 + e a: . 

(9) (i) y = \/(ax + b) ; (ii) t/ = ^/(alog x + b). 
(10) ?/ = (a cos x + b sin x + sin 2x) e x2 . 

(12) ?/ = x 2 2. (14) Z=-£. 

(17) (i) y = ae* 2 + fo-** 2 - sin x 2 . (Put 2 = x 2 .) 

(ii) ?/(l+x 2 )=a(l-x 2 ) + &x. (Putx = tanz.) 

(18) j^-2y = 2{l-z 2 ) ; ?/ = sin 2 x + 4 cos (V2 sin x + a). 

(19) y = acos{2(l+B)e-*} + 6sin{2(l+x)e^ x } + (i+*)e~" 



CHAPTEK VIII. 
Art. 83. 

(1) y = 2 + x + x 2 - ^x 4 - fvx 5 ; exact solution y = 2 + x + x 2 . 

(2) £/ = 2x - 2 log x - ^ (log x) 3 ; exact value y = x H 4 - . 

(3) ?/ = 2+x 2 + x 3 + t 3 tf x 5 + tV :k6 ; 

2 = 3x 2 + f x 4 + f x 5 + ^x 7 + /^x 8 . 

, (4) y = 5 + x + -^x 4 + -^x 6 + ^ 3 x 7 + T Vx 9 ; 

2 = 1 + ■?, X 3 + x 5 + I X 6 + ^X 8 + 5 V T X 9 + 2 * i x 11 . 

(5) 2/ has the same value as in Ex. 4. 

Art. 87. 
(1) 2-19. (2) 2-192. (3) (a) 4-12 ; (6) 4-118. 

(4) Errors 0-0018 ; 0-00017 ; 0-000013 ; 
Upper limits 0-0172 ; 0-00286 ; 0-000420. 

Art. 89. 

1-1678487; 1-16780250; 1-1678449. 



ANSWERS xi 

CHAPTER IX. 
Art. 95. 

(1) „w|i-* + _-...|»cobV3; v = a; i |l-|j+^-...|=8inV*. 

„L 8 8.11 „ 8.11.14 , I 
(4) ^ X H 1 "4TO x2 + 4.8(l + l)(2 + n) ^ 

1 6 \ 

4.8.12(l+w)(2 + n)(3 + w) a; + "'J' 
To get v from u change n into -n. If u is multiplied by 

the constant r-^, — , the product is called Bessel's function 

A n i (n + 1) 

of order n and is denoted by J n (x). 

Art. 96. 

(1) and (4), all values of x. (2) and (3), |x|< 1. 

Art. 97. 

f\ 2. 2.5, 2.5.10 . 1 

v = u log a; + { - 2x - x 2 - ^4-x 3 . . .}. 

(2) u = |l - - 2 x 2 + ^-p^ 4 _ 2 2 . 4 2 . 6 2a;6 + "" J ' 

t> = Mlogg+|^g 2 -^-^ i (l+^)g* + 22 ^^ 62 (l+| + ^)g B -...j. 

m is called Bessel's function of order zero and is denoted by 
J {x). 

(3) u={l-2x + ^x*-^x* + ...y, 

v = u\ogx+^2(2-i)x-~(2 + l-l)x^ + ^(2 + h + l-l)x 3 -...y 

if. 1.3 . 1.3.5.7 . 1.3.5.7.9.11 fi 1 

(4) MBa? »|l+ — a*+ ^ gg s*+ ^ 82 — 2l x 6 + ...|; 

M *} 

v = w log x + 2ar 1 t 2 (1 + i - 1) x 2 

P.D.E. p 2 



xii DIFFERENTIAL EQUATIONS 

Art. 98. 

i 

.r 8 
2 3 . 4 2 . 6 . 8 



(1) u-x 2 {- 2 2. 4 * 4 + 237T76 a;6 ~2 3 .4 2 .6. 

ro* 10 -] ; 

v = u\ogx + x-^l+^x 2 + ¥ -^-^-^^ 2 x 9 

31 8 | 

+ 2 2 A 2 .6 2 .8 2X "')' 

(2) u = x + 2x 2 + 3x* + ...=x(l-x)- 2 ; 

v = u log x + 1 + x + x 2 + . . . = u log x + (1 - a;) -1 . 

(3) w = {1.2a; 2 + 2.3x 3 + 3.4x 4 + ...}; 

v = ulogx + {-l+x + 3x 2 + 5x s + lx i + ...}. 

(4) M = {2a5 + 2a; 2 -a; 3 -a^ + |« 5 ...}; 

v = u log x + {1 - x - 5x 2 - x 3 + V"* 4 . . .}. 

Art. 99. 

(1)2/ = a {l - z 2 - ix 4 - is". . .} + ajx = a |l - |a log ^ J + a x x. 

(2) y^ji-^^ ^-^;; 1 )^^ -..,} . 

f (n-l)(n. + 2) (n-l)(n-3)(n + 2)(n + 4 ) . \ 

+ ai l 3! * +_ ~5T~ "7* 

[For solutions in powers of 1/x see No. 7 of the Miscellaneous 
Examples at the end of Chapter IX.] 

(3) y^{l-0^ + 1 7^ a?8 - 3.4.7.8.11.12 8l, + "j 

+ a ^-0^ 5 + 4T5V9 x9 - 4.5.8.9.12.13 xl3 + --} < 

(4) y = a {l - ^a 2 - T \x 3 + &x*...} + a x {x - \x* - ^x* + &*. . .}. 

Art. 100. 

(1) z^ + z^+{\-n 2 z 2 )y = 0. (2) y = ax 2 (l+2x). 

(3) y = x 2 (1 + 2b) {a + 6 j or 2 (1 + 2x)~ 2 e* dx}. 

(5) 2 e-and[ 2 e-log 2+2 2 {l-| ] (l+i) 2 + | I (l+^ + ^ 2 -...}], 

where z = \jx. 




ANSWERS xiii 

Miscellaneous Examples on Chapter IX. 

(1) u = x~l |l + - x + ^x 2 + ~x 3 + ...J; 

(13 9 _ 27 . ) 
V= \l! + 4! a; + 7!^ + lT! K+ -} ; 
i/l 3 9 _ 27 . I 
-^\2l + 6l X+ 8!^ + lT!^ + -T 

(2) M ,|l+J- a;+ -l^ a; 2 + i2 ^ 32 a *+„.}; ' 

t, = *dogz + 2|-pS-p r -^(l+-) a; 2 

1 2 .2 2 .3 2 \ 2 3/ '"/' 
w = u (log a;) 2 + 2 (v - u log a;) log x 



CHAPTER XI. 
Art. 113. 

(1) x/a = y/b = z', straight lines through the origin. 

(2) lx + my + nz = a ; x 2 + y 2 + z 2 = b; circles. 

(3) y = az; x 2 + y 2 + z 2 = bz; circles. 

(4) x 2 - y 2 — a ; x 2 -2; 2 = 6; the intersections of two families of rect- 

angular hyperbolic cylinders. 

(5) x-y = a (z-x) ; (x-y) 2 (x + y + z) = b. 

(6) x 2 + y 2 + z 2 = a ; y 2 - 2yz -z 2 = b; the intersections of a family of 

spheres with a family of rectangular hyperbolic cylinders. 

(7) y/(m* + n 2 ) . (8) The hyperboloid y 2 + z 2 - 2x 2 = 1 . 

(9) (x 2 + y 2 ) {k tsm-hj/x) 2 = z 2 r 2 . (10) l/x = l/y + 1/2 = \(z + 2. 

Art. 114. 

(1) y-3x = a; 5z + t&n(y-3x)=be^ x . 

(2) y + x = a; log {z 2 + (y + x) 2 } - 2x = b. 

(3) xy = a; (z 2 + xij) 2 -x i = b. (4) y = ax\ \og(z-2x/y)-x = b. 

Art. 116. 

(1) x 2 + y 2 + z 2 = c 2 ; spheres with the origin as centre. 

(2) x 2 + y 2 + z 2 = ex ; spheres with centres on the axis of x. 

(3) xyz — <?. 



XIV DIFFERENTIAL EQUATIONS 

(4) yz+zx + xy = c? ; similar conicoids with the origin as centre. 

(5) x-cy=y log z. 

(6) x 2 + 2yz + 2z 2 = c 2 ; similar conicoids with the origin as centre. 

Art. 117. 

(1) y — cx log z. (2) x 2 y = cz&. (3) (x + y + z 2 )e x2 = c. 

(4) y{x + z) = c(y + z). (5) {y + z)/x + (x + z)ly = c. 
(6) w/-mz = c(wa?-fe). The common line is x\l = y\m=z\ , n,. 

Art. 120. 

(3) z = ce 2 *. (4) z 2 z + 4=0. 

Miscellaneous Examples on Chapter XI. 

(1) y = ax; z 2 -xy = b. (2) x^y % z — a\ x z + y z = bx 2 y 2 . 

(3) y + z = ae x ; y 2 -z 2 = b. (4) ?/ = sina; + cz/(l + z 2 ). 

(5) x 2 + a:?/ 2 + x 2 z = £ + c. (6) f(y) = ky; x k = cy z . 

(8) <fo/a; = %/2?/ = dz/3z. (9) y + z = 3e*- 3 ; y 2 - z 2 = 3. 

(10) (i) x 2 + y 2 + z 2 = c(x + y + z) ; (ii) x 2 - xy + y 2 = cz ; 

(iii) y 2 -yz-xz = cz 2 . 
(14) xt/ = ce z sin w. 

CHAPTEE XII. 
Art. 123. 

(1) < p(x/z ! yfz)=0. (2) (j>(lx + m,y + nz,x 2 + y 2 + z 2 )=0. 

(3) 0{*//z, (x 2 + y 2 + z 2 )jz} = 0. (4)0 (x 2 - y 2 , x 2 - z 2 ) = Q. 

(5) <p{(x -y) 2 (x + y + z),(x- y)l(z - x)} = 0. 

(6) (p{x 2 + y 2 + z 2 , y 2 - 2yz - z 2 } = 0. 

(7) <f>[y-3x, e~ 5a: {5z-tan(«/-3a;)}]=0. 

(8) </>{y + x, log (z 2 + y 2 + 2yx + x 2 ) - 2x) = 0. 

(9) y 2 = ixz. (10) a(x 2 -y 2 )+b(x 2 -z 2 )+c = 0. 
(12) (j>(x 2 + y 2 , z) =0 ; surfaces of revolution about the axis of z. 

Art. 126. 

(1) (fi{z + X x ,X 1 J fX 2 ,X- i + X 3 )=0. 

(2) 0(z, xfrf 1 , xfxf 1 , a; 1 %r 1 )=°- 

(3) (p(z-x 1 x 2 , a^ + Xa + Zg, x 2 x 3 )=0. 

(4) 0(2z + a; 1 2 , x x 2 -x 2 2 , x x 2 -x 3 2 )=0. 

(5) (p(4\/ z ~ x z 2 > 2x 3 -x 2 2 , 2x 2 -a; 1 2 )=0; special integral z = 0. 

(6) <p{z-3x v z-3x 2 , z + 6\ / (z-x 1 -x 2 -x 3 )}=0; special integral 

Z == X-\~tX 2 ~tX 3 . 



ANSWERS XV 

Art. 129. 

(1) z = (2b 2 + l)x + by + c. (2) z = a;cosa + y8in a + c. 

(3) z = ax + y\oga + c. (4) z = a s x + a~ 2 y + c. 

(5) 2 = 2x sec a + 2y tan a + c. (6) z = a;(l + a) + ?/(l + l/a) + c. 

Art. 130. 

(1) az = (a + ay + bf. (2) z=± cosh {(a; + ay + 6)^(1 + « 2 )}- 

(3) z 2 -a 2 = {x + ay + b) 2 , or 2 = 6. (4) z 2 (l+a 3 )=S{x + ay + bf. 
(5) (2 + a)e a! + a 2/ = 6. (6) 2 = 6e ax+a X 

Art. 131. 

(1) 3z = 2(x + a)* + 3ay + 3b. (2) 2az = a 2 x 2 + y 2 + 2ab. 

(3) az = ax 2 + a 2 x + e a v + ab. (4) (2z-ay 2 -2b) 2 = l6ax. 

(5) 2 = a(e x + e 2/ ) + 6. (6) az = a 2 x + a sin x + sin */ + a6. 

Art. 133. 

(1) 2 = - 2 - log xy. (2) 3z = xy-x 2 - y 2 . (3) 82 3 = - 27x 2 y 2 . 

{±)zx=-y. (5)2 = 0. (6) 2 2 = 1. (7)2 = 0. 

Art. 136. 
(1) 4«= -y\ 

(4) A particular case of the general integral, representing the surface 

generated by characteristics passing through the point (0, -1,0). 

Miscellaneous Examples on Chapter XII. 

(1 ) 2 = ax + by - a 2 b ; singular integral z 2 = x 2 y. 

(2) zx = ax + by- a 2 b ; singular integral z 2 = y. 

(3) <j>{xy,(* 2 + xy) 2 -x*}=0. 

(4) 2 = 3z 3 - 3ax 2 + a 2 x + 2?/ 4 - lay 3 + 3ahj 2 - a 3 y + b. 

(5) 2 = ax x + b log x 2 + (a 2 + 26) x 3 _1 + c. 

(6) z^^^ + x^x^x^-x^}. 

(7) 3a(x + at/ + 6) = (l +a 3 )log2, or 2 = 6. 2=0 is included in z = 6, but 

it is also a singular integral. 

(8) z(l+a 2 + b 2 ) = (x 1 + ax 2 + bx 3 + c) 2 . 

(9) <p(z-e 4a; i, 2-e 4 ^,2-e 4 ^)=0. (10) z = ax- (2 + 3« + hr)y + b. 
(11) 2 2 = az 2 -(2 + 3a + ia 2 )2/ 2 + 6. (12) 2 2 = (1 + a 2 )x 2 + aif + b. 

(13) z = a tan (£ + a?/ + 6), or 2 = 6. 2 = is a singular integral, but it is 

also included in z = 6. 

(14) 2 2 = ax 2 + by 2 - 3a 3 + 6 2 . Singular integral z 2 = ± 2x 3 /9 - i/fi. 

(15) z=x + y-l±2y/{{x-l){y-l)}. (1(3) z 2 -xy=c. 



XVI DIFFERENTIAL EQUATIONS 

(17) <j>{z/x, z/y)=0 ; cones with the origin as vertex. 

(18) x 2 + y 2 + z 2 = 2xc6sa + 2yB,ma + c; spheres with centres on the 

given circle. 

(19) xyz = c. (This is the singular integral. The complete integral 

gives the tangent planes.) 

(20) The differential equation (z - px - qy) (1 - 1/p - 1/q) = has no 

singular integral, and the complete integral represents planes. 



CHAPTER XIII. 
Art. 139. 

(1) y 2 {{x-a) 2 + y 2 + 2z} = b. (2) z 2 - 2ax + a 2 y 2 + b. 

(3) z = ax + bey(y + a)~ a . (4) z 2 = 2(a 2 + l)x 2 + 2ay + b. 

(5) z = ax + 3a 2 y + b. (6) (z 2 + a 2 ) z = 9(x + ay + b) 2 . 

(7) z = x 3 + ax + %(y + afi 2 + b. (8) z = ax + by + a 2 + b 2 . 

Art. 141. 

(l)z = a 1 x 1 + a 2 x 2 + (1 - a>\ - « 2 2 ) x z + a z . 

(2) z = a 1 x 1 + a 2 x 2 ± sin -1 (c^ag,^) + a z . 

(3) z = a x log x x + a 2 log x 2 ± x z -\/{a x + a 2 ) + a z . 

(4) 2z = a x x x 2 + a 2 x 2 2 + a z x z 2 - 2 {a x a 2 a z ) l l z log a: 4 + a 4 . 

(5) 2{a x a 2 a z ) x < z log z = a^ 2 + a 2 x 2 2 + a z x z 2 + 1. 

(6) ia x z = 4a x 2 log x z 4- 2a 1 a 2 (x 1 - # 2 ) - (x x + x 2 ) 2 + ^a x a z . 

(7) (1 + «!a 2 ) log z = (a x + a 2 ) (x x + a x x 2 + a 2 x z + a z ). 

(8) z = -(a x + a 2 )x x + (2a x -a 2 )x 2 + ( -a x + 2a 2 )x z 

- ^{x x 2 + x 2 2 + x 3 2 ) ± § {x x + x 2 + x z - 2a x 2 + 2a x a 2 - 2a 2 2 } 3 / 2 + a 3 . 

Art. 142. 

(1) z = =h (x x + x 2 ) 2 + log x z + a. (2) No common integral. 

(3) z = x x 2 + x 2 2 + x z 2 + a, or z = x x 2 + 2x 2 x z + a. 

(4) z = a (x x + 2x 2 ) + b log x z + 2ab log x 4 + c. 

(5) z = a(3x x + x 2 3 -x z 3 ) + b. (6) No common integral. 

(7) z = a(x 1 -cc 4 ) + fe(a; 2 -a;3) + c J or z = a(x 1 -2x 2 ) + b(2x z -x 4 ) + c. 

(8) Z = 0(3.2! + z 2 3 -z 3 3 ). 

(9) z = (f>{x x -x /x , x 2 -x 3 ), or z = 0(£ 1 -2# 2 , 2x 3 -x 4 ). 

Miscellaneous Examples on Chapter XIII. 

(1) z 2 = a x log x x - a x a 2 log £ 2 + a 2 log£3 + a 3 . 

(2) No common integral. 

(3) z = a x log X! + a 2 x 2 + (a x + a 2 ) x z ± \/{ a i ( a i + % a z) x i 3 l + a 3- 



ANSWERS Xv ij 

(4) = a x log x y + a& 2 + (a 2 + a 2 ) x 3 ± y/{a x (a x + 2a 2 ) z 3 } + 1 . 

(5) 2logz = c±(x* + x* + x 3 z ). (6) z 3 = x* + x.* + x 3 3 + c. 

(7) 42 + aj 1 2 + x 2 2 + x 3 2 = 0. (10) z = <p( Xl x 2 , x 2 + x 3 + x A , x A x b ). 

(11) (iii) 3z = x 1 3 -3x 1 x 2 +c. 



CHAPTER XIV. 
Art. 144. 
z = x* + xf(y) + F(y). (2) z - log x log y +f(x) + F(y). ^ 

z=-^sinxy + yf(x) + F(x). (4) z - xy +/(y) log x + F(y). 

z = sin(x + y) + -f(x) + F(y). (6) z= -xy+f(x) + e*yF(x). 

z = (x 2 + y 2 ) 2 -l. ' (8) z = ^ 2 + 2x?/ + 2</ + ua; 2 + &;r + c. 

2 = (x 2 + i/ 2 ) 2 . (10) 2 = ^ + |/(l-^) 

Art. 145. 
«- J P 1 (y+«) + JF , 2 (y + 2a;) + f 8 (y + 3a;). 

«=/(y-2z) + JF(2y-a;). (3) z=f(y + x) + F(y-x). 

The conicoid 4a; 2 - 8xy + y 2 + 8x-4:y + z=0. 

Art. 146. 

z =f(2y - 3a?) + xF(2y - 3x). (2) 2 =f{5y + ix) + xF(5y + ix). 

z=f(y + 2x)+xF(y + 2x) + <j>(y). (4) z(2x + y)=3x. 

Art. 147. 

z = x 4 + 2a% +/(?/ + a;) + «/(y + x). 

z = 2{f-x 3 ) +f{y + 2x) + F {2y + x). (3) V = - 27ra: 2 ^ 2 . 

Art. 148. 

z = e x + 2 !>+f(y + x)+xF(y + x). 

z = x 2 (3a; + y) +f(y + 3x) + xF(y + 3x). 

z = - x 2 cos {2x + y) +f{y + 2x) + xF(y + 2x) + <p ( y). 

z = xe x -y+f{y -x) + F(2y + 3x). 

V = (x + yf +f(y + ix) + F(y - ix). 

z = 2x l log(cc + 2y) + f(2y + x) + xF(2y + x) 

Art. 149. 

z = x sin y+f(y-x) + xF(y -x). 
z = x* + 2x*y +f{y + 5x) + F(y - 3.x). 
z = sin x - y cos x + f(y - 3x) + F(y + 2x). 
z = sin xy+f(y + 2x) + F(y - x). 



xviii DIFFERENTIAL EQUATIONS 

(5) z = %ta,nxta,ny+f(y + x) + F(y-x). 

(6) y = x log t + 1 log x +f(t + 2x) + F(t- 2x). 

Art. 150. 

(1) z=f{x) + F(y)+e? x <t>(y + 2x). 

(2) z = er*{f(y -x) + xF(y- x)}. (3) 7 - 2Ae hlx+M) . 

(4) z=f(y + x) + e-*F{y-x). (5) z = ^Ae h{x+hv) + y ZB^ x+7k >>\ 

(6) 3 = 2^e n(a:C03a -H' 8in ' l ». (7) z = e x {f(y + 2x) + I l Ae! civ+ikx )} 

(8) « = l+e- a! {(2/-a;) 2 -l}. 

Art. 151. 

(1) z = \e 2x ~v + e x f(y + x) + e 2x F(y + x). 

(2) z = l+x-y-xy + e x f{y) + e-vF(x). 

(3) z = sV( sin (x - 3y) + 9 cos (x - 3y) }% ZAe^+^l 

(4) z = x+f(y)+e- x F(y + x). (5) 2/ = |xe x+2; + 2^e j:seCa + rtana . 
(6) z = e 2x {x 2 tan (t/ + 3z) +»/(«/ + 3x) + F(y + 3x)}. 

Art. 152. 

(1) y 2 r-2ys + t=p + 6y. (2) pt-qs=pq 3 . 

(3) r + 3s + * + (rt-s 2 ) = l. 

(4) pq(r-t)-(p 2 - q 2 ) s + (py - qx) (rt - s 2 ) = 0. 

(5) 2^7 , + gtf-2^(rt-s 2 ) = l. (6) qr + (zq-p)s-zpt = 0. 

Art. 154. 

(1) z =f{y + sin a:) + F(y - sin x). (2) z =/(x + ?/) + ^(xy). 

(3) y-\p-{x + y + z) = <f>(x), or z=f(x) + F(x + y + z). 

(4) 2 =/(£ + tan !/) + F(x - tan */). (5) z =f(x 2 + y 2 ) + F{y/x) + xy. 

(6) y=f{x + y + z)+xF{x + y+z). (7) 3z = 4^ - x 2 xf - 6 log y - 3. 

Art. 157. 

(1) p + a;-2t,=/(g>-2a; + 3y) ; A = -£. 

(2) p-x=f(q-y); A = a>. (3) p-e x =f(q-2y) ; A = a>. 

(4) p-y=f{q + x); p + y = F(q-x); \=±1. 

(5) p-y^f{q~2x) ; p-2y = F(q-x) ; A = -1 or -£. 

(6) px-y=f{qy-x); \=-xox -y. 

(7) z^ - z =/(z£ - 1/) ; A = z/^ 2 or z/y 2 . 

Art. 158. 

(1) z = az + by - \x 2 + 2xy - f y 2 + c ; 

z = |a; 2 ( 1 + 3m 2 ) + (2 + 3m) xy + nx + <j>(y + mx) 
= 2xy - \ (x 2 + 3y 2 ) + nx + \js(y + mx). 






ANSWERS 



XIX 



(2) « = f(a* + y*)+oa; + &y + c; z^^ + y^ + nx + y^iy + mx). 

(3) z = e x + y 2 + ax + by + c; z = e x + y 2 + nx + yf,(y + mx). 

(4) & = i(a-/3); 2/ = HV / (^)-0'(«)}; z = ™J + \{<t>{a)-y},(p)} + Py. 

(5) a: = £- a ; y = <t>'{a)-\[r'(l3); z = xy-<f>( a )+yj,{P)+Py. 

(6) z + y/m + mx-n\ogx = <p(x m y) ; the other method fails. 

(7) z 2 = a? + y 2 + 2aa; + 2oy + c; z 2 = x 2 + y 2 + 2nx + \J,(y + mx). 

(8) 2z = y 2 -a; 2 . 

Miscellaneous Examples on Chapter XIV. 

(1) z = x 2 y 2 + xf(y) + F(y). (2) z = e*+v+f(x) + F(y). 

(3) yz == y log y -/(a) + yF{x). 

(4) z=/(cc + ?/)+a;i , (a; + ?/)-sin(2a; + 3?/). 

(5) 2=/(y + loga;)+*JP(y + loga;). (6) z = x + y+f{xy) + F(x 2 y). 

(7) 3 = log(* + y) ./(a; 2 - 1/ 2 ) + *V - if). 

(8) 42 = 6a;y - 3a; 2 - 5y 2 + lax + 46?/ + c ; 

4z = 6a;?/ - 3a; 2 - 5y 2 + 2nx + 2\js (y + mx). 
. (9) 3z = 3c±2{x + a) 3 l' 2 ±2(y + b) 3 ' 2 . 

(10) mz + sin y + m 2 sin x - mnx = m<f>(y + mx). 

(11) 2a; = «-/3; fy = ^'(/3) - 0'(«) J 

22 = 3a; 2 - 6^ - 7y 2 + </>(«)- -f (/3) + 2fty. 

(12) z = x 3 + ?/ 3 + (a; + ?/ + l) 2 . (13) z = x 2 -xy + f. 
(20) px + qy=f(p 2 + q 2 ); py -qx= F(q/p). 

Miscellaneous Examples on the Whole Book. 
? (1) {x 2 -y 2 ) 2 = cxy. (2) ?/ = a; 2 + ce- r2 . 

(3) 2 sec a; sec ?/ = a; + sin a: cos a; + c. (4) (a;?/ + c) 2 = 4(a; 2 + ?/)(?/ 2 -ca;). 
(5) 1 +a;?/ = y/(c + sin- 1 x)v / (l -a; 2 ). (6) y = (^ - ]x) cos 2x + JBsin 2a;. 

a; 2 6a; 28 1 

( 7 ) !/ = ^-^ + ] 2? + ^^ a: (sin2a;-cos2a;) + Je- a: + 5e a: cos(2a; + u). 

(8) y = A + Bx + Cx log x + log x + |a;(log x) 2 + la; 2 . 

(9) y + sec x = c tan a;. 

(10) a; = Ae lt + Be" 2 ' - f (cos « - sin <) ; y = ,4e 2( - 3/fr- 2 < - « cos t. 

(11) a; 2 / 3 = (y-l) 2 / 3 + c. (12) y = acosvc(b-x). 

(13) y = L4 + Bx + X J sin 2x + ( J0 + Fx - J cos 2a;. 

(14) 2xy = 3x 2 + c. (15) z + xy = c(x + y -xy). 
(16-) a; 3 + y 3 + 2 3 = ca;v/z. (17) 2 =/(.n/) - Jx 2 - -J y 2 . 

(18) (x - y) ^-^^) =/{(» - 3y + z)/(x - yf). 

(19) (2 + a;) 2 = (2 + 2y)/0//x-). 

(20) z = ax + by + a 2 + b 2 ; singular integral iz + x 2 + ij- = 0. 

(21) z = e*f(x-y) + F(y). 



XX DIFFERENTIAL EQUATIONS 

(22) z = aa* 2 + % + 4a 2 ; singular integral 1 62 + cc 4 =0. 

(23) z=f(x + y) + F(x-y) + b(x* + f). 

(24) z = xf(y) + yF(x). (25) cz = (x + a)(y + b). 
(26) z = lxy+f(y/x) + xF(yfx). (27) z=f{z + x) + F(z + y). 

(28) y(x + c) = c 2 x; singular solutions y = and «/ + 4a; 2 = 0. 

(29) atf = {x + bf. (30) 2/ = ^cos(|^) + 5sin(^). 

(31) r 5 + «/ 2 + z 2 = 2(a;cosa + ?/sina + c). (32) y = e x -\ e ix + \^ x . 

(33) a; = e~ Kt (a cos X£ 4- b sin XZ) + C cos (pt - a), 

where C = .4/ ^/{(/c 2 + X 2 - flf + ±K 2 p 2 }, tana - 2 K pJ( K 2 + X 2 - 2> 2 ), 
and a and 6 are arbitrary constants. 

(34) y = A cos (sin x) + B sin (sin x). 

(35) (i) i^ = ^log(r + 2) + 5; 

(ii) <f> - A [ e-^ 4a ' 2 d£ + 5 ; |^ = ~ e^l***. 

(36) F = A{\ + f (3z 2 - r 2 ) + ^V (35z 4 - 30z 2 r 2 + 3r% 

where r 2 = x 2 + y 2 + z 2 . 



l+- + 7T-I + £7-*+'» COsh< 

a 4! a 4 5! a 5 / 



«(£ 



X 3 X 6 X \ . , 

2 + K7— 3 + ^7— 6 + srj— 7 + -- Jsinn*. 
12 31a* 1 6 ! a b 7 ! a 7 / 



(41) y-x = c(xy -l)e~ x . 

(42) ?/ = (l+a;) a - 6 (l-cc) a +V + ^ f (l+^) _0+6-1 (l-^)" a " 6-1 ^}- 

If 2a is an integer, the integral can be evaluated by 
putting z = ( 1 + x)/( 1 - x) . 

(43) (i) y = (l-x 2 ){A + B\ogx); (ii) y = {l -x 2 )(x + A + Biogx). 

(44) (1 - x 2 ) y = (a + b e-* 2 oa;) e^ x ' 2 - [Put log y = (w - 1 P) dx. u = x is 

a solution of the differential equation in u.] 
<M fM-l (2»-2)x 2 (2H -2)(2n-4)(2n-6) x 4 . 
l«);/w (2n-l)2! (2«-l)(2n-2)(2n-3) 4! *" • 

(2 w-2)(2w-4) a 3 
^ (a;, ~" (2n-l)(2n-2)3! + "" 
(46) */ = ,4a- 5 + J5r } + #(a; 2 + l), replacing C/6 by £. 



m--^**^© 1 



c{c + 2(6 + l)}{c + 4(6 + 3)} /x\ 6 

+ 6! ' W + 

5 
+ .., 



Va/ 3! Va/ 5! Va/ 

both converge within the circle \x\ < \a\. 



ANSWERS xx i 

(50) q j i ^- i must be a function of x alone ; x*y - ax 2 y 2 = c. 

(51) x* + y 2 + 2bxy = 2ax. 

(52) uve w = a I vV dx + 6, where t; = Q/P and w = It; dx. 

(53) Pwcot(nx + a)+Q = n 2 . 

(54) y(l-x)=A(3 - 2x) e 2x +B(l-2x) e~ 2x . 

(56) x 3 + yz = c(y + z). 

(57) t/ = ^e- 2a5 + e a; (5cosa;V3 + Csina;V 3 ) 

+ t x2 +2riTir e_2a; {157x(6 cos a; + 11 sin a;) 

+ 3 (783 cos a; -56 sin x)}. 

(58) ^ = (3 + 4a; 2 ){^ + £f(3 + 4a; 2 )- 2 e-^ 2 ^}. 

(59) z*(x + y)*{x 2 + y 2 + z 2 )=c{x 2 + y 2 -z 2 ). (60) xz = c(y + z). 

(62) (1) Put «= rr-5~; (") y~K=-r -• 

* ua 2 (x) dx K ' v 2 l-ctanz 

[See Ex. 41 for method.] 

(65) If a particle P moves so that its velocity is proportional to the 

radius vector OP and is perpendicular to OP and also to a 

fixed line OK, then it will describe with constant speed a 

circle of which OK is the axis. 

(67) r 2 sin 2(0 + a) = l ; singular solution r*=*=l. 

(68) y 2 - x 2 = ex + 2a 2 + a\/(4a 2 - c 2 ) ; singular solution y 2 - x 2 = ± 2a y. 

(70) ia(y - c) = (x - c) 2 ; singular solution «/ = x -a. 

(71) x + a = c cos (j) + c log taji |0. (72) a cos + b cos 0' = k. 

(74) 2c?/ = (a; + c) 2 ; singular solution y(y -2x)=0. 

(75) a;+j9«/ + ajj 2 = 0; (2/ + a^)v / (p 2 + l) = c + a sinh -1 ^, 
x V(P 2 + l)+p(c + a sinh -1 ^) = 0. 

There is no singular solution. The ^-discriminant y 2 = iax 
represents the cusp-locus of the involutes. 

(77) y = ax,z = b + V(x 2 + y 2 ); * = V(* 2 + // 2 ) +/(.<//•')• 

The subsidiary integrals represent a family of planes through 
the axis of z and a family of paraboloids of revolution with 
the axis of z as axis ; the general integral represents a family 
of surfaces each of which contains an infinite number of the 
parabolas in which the planes and paraboloids intersect. 

(78) x 2 + y 2 + z 2 =f{x 2 + y 2 + (x + y) 2 } ; a; 2 + if + z 2 = c 2 ; z 2 = xy. 

(79) (2x-y)' 7 = c 5 z{x + 2y). 

(80) (ax - by)j(z + c) =f{(ax + by)j(z - c)}. 



xxii DIFFERENTIAL EQUATIONS 

(81) (i) I=EIR + Ae- Rt ' L ; (ii) A = I -E/R; (iii) I=EjR. 

(82) I = a cos {pt - e) + Ae - Rl ' L , where a = E/^R 2 + L 2 p 2 ), tan e - Lp/R, 

and A is arbitrary. 

(83) Q = a sin (p< - e), where tan e = (C-Lp 2 - l)jpCR and 
a - EC/V{(CLp* - l) 2 +p 2 C 2 R 2 }. 

(85) z = 4 cos {t- a) + B cos (3t - ft) ; y = 2A cos (t-a)-5B cos(St-ft) 

(86) a and b are the roots of \ 2 {LN - M 2 ) + \{RN + LS) + RS = 0. 

(91) x = A cos {pt -a) + B cos (gtf -fi),y = A sin (p* - a) - B sin (<^ - /3), 

where 2p = y^c 2 + k 2 ) +K,2q = V(±<? + * 2 ) - * • 

(92) -^ + (a + 6) -=- + abz = a&c. 

(93) 2? = \/( w2 _ V 2 ) makes the amplitude of the particular integral a 

maximum, provided 2/j 2 does not exceed n 2 . 

(94) x = ^e~ w cos {pt - e), where ^ = \/{n 2 - h 2 ). 

(97) <j> = \ Fa 3 r- 2 cos 0. (98) y sin (pfr/c) = J. sin (yx/c) (cos ^ + a). 

(100) = Ccosh m{y + h) cos {mx-nt). 
(115) (vi) u.-4(-2)"+B(-l)«>; 

(vm) u x = 2 X (P cos — +Q sm — J , 

(x) Ma! = 4(-9)* + B+^. 



INDEX 



(The numbers refer to the pages.) 



Ampere, xvi, 183. 

Angstrom's determination of diffusivity, 

58. 
Approximate methods, 5, 94, 209. 
Arbitrary constants, 2, 50, 126, 127, 

214. 
Arbitrary functions, 49, 137, 147, 172. 
Asymptotic series, 213. 
Auxiliary equation, xv, 26, 174, 216. 

Bar, vibrating, 190. 

Bateman, 194. 

Bernoulli, xv, 12, 18. 

Bernoulli's equation, 18. 

Bessel, 110. 

Bessel's equation, 114, 116, 118, 120, 

215. 
Boole, xv. 

Boundary conditions, 53, 56. 
Briot and Bouquet, xvi. 
Brodetsky's graphical method, vi, 5. 
Bromwich, 209. 

Cauchy, xvi, 121, 124. 

Cayley, xv. 

c-discriminant, 67, 155. 

Change of variables, 40, 61, 79, 85, 91, 

93, 119, 120, 164. 
Characteristics, 6, 97, 158. 
Charpit, xvi, 162. 
Charpit's method, 162. 
Chemistry, 206. 
Chrystal, xvi, 150. 
Clairaut, xv, 76. 
Clairaut's form, 76, 79. 
Common primitive, 10. 
Complementary function, 29, 87, 175, 

216. 
Complete integral, 153. 
Complete primitive, 4. 
Conditions of integrability, 139, 144, 

191, 193 



Conduction of heat, 52, 53, 57, 58, 59, 

60, 212. 
Confocal conies, 23, 79. 
Conjugate functions, 24, 189. 
Constant coefficients, xv, 25, 49, 173, 

178, 212, 214, 216. 
Constants, arbitrary, 2, 50, 126, 127, 

214. 
Convergence, xvi, 112, 124. 
Corpuscle, path of a, 48. 
Cross-ratio, 201. 
Cusp-locus, 68, 73. 

D'Alembert, xv, 25, 44, 49. 
Darboux, xvi. 

Definite Integrals, solution by, 212, 213. 
Degree, 2. 

Depression of order, 81. 
Developable surface, 189. 
Difference equations, 216. 
Difficulties, special, of partial differen- 
tial equations, 51. 
Diffusion of salt, 60. 
Discrinnnant, 67, 71, 155. 
Duality, 160, 161, 18!), 210. 
Dynamics, 2, 24, 28, 36, 46, 47, 50, 61, 

*85, 86, 190, 204, 205, 206, 207, 208, 

200,210, 211. 

Earth, age of, 60, 212. 

Einstein, 209. 

Electricity, 24, 29, 4(i, 48, 58, 59, 134, 

203, 204, 205, 206. 
Elimination, 2, 4!), 50, 179. 
Envelope, 60, 71, 146, 155. 
Equivalence, 92. 
Kuler, xv, 12, 25, 49. 
Exact equations, 12, 23, 91, 191. 
Existence theorems, 121. 214. 

Factorisation of the operator, 86. 
Falling body, 24, 86. 



XXIV 



DIFFERENTIAL EQUATIONS 



(The numbers refer to the pages.) 



Falling ohain, 208. 

Finite differences, 215, 216. 

First order and first degree, ordinary, 

12, 133 ; partial, 147, 151. 
First order but higher degree, ordinary, 

62, 65 ; partial, 153, 162, 165. 
Fontaine, xv. 
Forsyth, 150, 194. 
Foucault's pendulum, 209. 
Fourier, 54. 
Fourier's integral, 60. 
Fourier's series, 54. 
Frobenius, xvi, 109. 
Frobenius' method, 109, 127. 
Fuchs, xvi. 
Functions, arbitrary, 49, 137, 147, 172. 

Gauss, 110. 

General integral, xvi, 137, 147, 149, 157. 

General solution, 4. 

Geometry, 5, 19, 65, 133, 137, 146, 173, 

188, 189. 
Goursat, xvi, 172, 194. 
Graphical methods, 5, 8. 
Groups, xvi, 120, 194. 

Hamilton's equations, 210. 

Heat, 52, 53, 57, 58, 59, 60, 212. 

Heaviside, 58, 61. 

Heun, 94. 

Heun's numerical method, 104. 

Hill, M. J. M., vi, xv, xvi, 65, 150, 155, 

192, 194. 
Homogeneous equations, xv, 14, 40, 44, 

83, 144, 171, 173, 213. 
Homogeneous linear equations, 40, 44, 

171, 173. 
Hydrodynamics, 208. 
Hypergeometric equation, 119, 120. 
Hypergeometric series, 92, 119. 

Indicial equation, 109, 111. 
Initial conditions, 4, 28, 53. 
Inspection, integration by, 12, 172. 
Integrating factor, xv, 13, 17, 22, 23, 91, 

199. 
Integrability, 139, 144, 191, 193. 
Integral equation, 96. 
Intermediate integral, 181. 
Invariant, 92. 

Jacobi, xvi, 165. 

Jacobi's Last Multiplier, 211. 

Jacobi's method, 165, 193, 210. 

Kelvin, 58, GO, 212. 
Klein, xvi. 



Kutta, 94, 104, 108. 

Kutta's numerical method, 104. 

Lagrange, xv, 49, 81, 162. 
Lagrange's dynamical equations, 210. 
Lagrange's linear partial differential 

equation, xvi, 147, 151, 158, 192. 
Laplace, xvi. 
Laplace's equation, 51, 189, 190, 196, 

197, 213. 
Last multiplier, 211. 
Laws of algebra, 30. 
Legendre, 110. 

Legendre's equation, 117, 120. 
Leibniz, xv. 
Lie, v, xvi, 194. 

Linear difference equations, 216. 
Linear equations (ordinary), of the 

first order, 16, 214 ; of the second 

order, 86, 87, 88, 109, 127, 214, 215; 

with constant c -efficients, xv, 25, 

214. 
Linear equations (partial), of the first 

order, xvi, 50, 147, 151, 158, 192; 

with constant coefficients, 49, 173, 

178, 212. 
Linearly independent integrals, 216. 
Lines of force, 24, 134. 
Lobatto, xv. 

Maxwell's equations, 59. 
Mechanics, see Dynamics. 
Membrane, vibrating, 190. 
Monge, xvi, 172. 
Monge's method, 181, 183. 
Multipliers, 135, 210, 211. 

Newton, xv. 
Node-locus, 68. 

Non-integrable equations, 142. 
Normal form, 91, 92. 
Normal modes of vibration, 204, 206. 
Number of linearly independent inte- 
grals, 216. 
Numerical approximation, 94. 

One integral used to find another, 87. 

136. 
Operator D, 30, 44, 86, 174, 214. 
Operator 0, 44. 
Orbits, planetary, 86, 209. 
Order, 2. 
Orthogonal trajectories, xv, 20, 23, 138, 

189. 
Oscillations, xv, 2, 28, 29, 36, 46, 47, 

48, 50, 61, 190, 203, 204, 205, 206, 

207. 



INDEX 



XXV 



(The number a refer to the pages.) 



Page, 194. 

Particular integral, xv, 4, 29, 33, 44, 87, 
175, 178, 216. 

p-discriminant, 71, 155. 

Pendulum, 28, 206, 207, 209. 

Perihelion of Mercury, 209. 

Physics, see Conduction of heat, Cor- 
puscle, Diffusion, Dynamics, Electri- 
city, Hydrodynamics, Potential, Ra- 
dium, Resonance, Telephone, Vaporisa- 
tion, and Vibrations. 

Picard, xvi, 94, 121. 

Picard's method, xvi, 94, 122. 

Poincare, xvi. 

Poisson's bracket expression (F,Fj), 166. 

Poisson's method, 189. 

Potential, 134, 190. 

Power series, xv, xvi, 4, 109, 124. 

Primitive, 4. 

Radium, 24. 
Reduction of order, 81. 
Regular integrals, 1 10, 1 18. 
Resonance, 37, 46, 205. 
Riccati, 110. 

Riccati's equation, 119, 201. 
Riemann, vi, 194. 
Runge, xvi, 94, 99, 100. 
Runge's numerical method, 99. 

Schwarz, xvi, 92. 

Schwarzian derivative, 92. 

Schlesinger, 194. 

Second integral found by using a first, 

87, 136. 
Separation of the variables, xv, 13. 
Series, solution in, xv, xvi, 4, 109, 124. 
Shaft, rotating, 47. 

Simple harmonic motion, 2, 85, 204, 206. 
Simultaneous equations, 42, 59, 133, 

168, 171, 214. 
Singular integral, 155. 



Singular point, 7. 

Singular solution, xv, 4, 65. 

Solid geometry, 133, 137, 146, 173, 188, 

189. 
Solving for p, x, or y, 62. 
Special integral, 137, 150, 192. 
Standard forms, 153. 
String, vibrating, xv, 50, 61, 190, 208. 
Subsidiary equations, 147, 164, 166. 
Substitutions, 40, 61, 79, 85, 91, 93, 119, 

120, 164. 
Symbolical methods, xv, 33, 44, 45, 61, 

175, 178, 214. 

Tac-locus, 72. 

Taylor, xv. 

Telephone, 58. 

Todd, 213. 

Total differential equations, 137. 

Transformations, 40, 61, 79, 85, 91, 93, 

119, 120, 164. 
Transformer, electrical, 48. 

Vaporisation, 24. 
Variation of parameters, 88, 93. 
Vibrations, xv, 2, 28, 29, 36, 46, 47, 48, 
50, 61, 190, 203, 204, 205, 206, 207. 

Wada, xvi. 5, 8, 9. 
Weber, 194. 

Whittaker and Watson, 214. 
Whittaker's solution of Laplace's equa- 
tion, 51, 213. 
Wronski, 215. 
Wronskian, 215. 

x absent, 82. 

y absent, 82. 

Zeemann effect, 206. 



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371 An elementary treatise on 
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